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David Howells108b42b2006-03-31 16:00:29 +01001 ============================
2 LINUX KERNEL MEMORY BARRIERS
3 ============================
4
5By: David Howells <dhowells@redhat.com>
David Howells90fddab2010-03-24 09:43:00 +00006 Paul E. McKenney <paulmck@linux.vnet.ibm.com>
David Howells108b42b2006-03-31 16:00:29 +01007
8Contents:
9
10 (*) Abstract memory access model.
11
12 - Device operations.
13 - Guarantees.
14
15 (*) What are memory barriers?
16
17 - Varieties of memory barrier.
18 - What may not be assumed about memory barriers?
19 - Data dependency barriers.
20 - Control dependencies.
21 - SMP barrier pairing.
22 - Examples of memory barrier sequences.
David Howells670bd952006-06-10 09:54:12 -070023 - Read memory barriers vs load speculation.
Paul E. McKenney241e6662011-02-10 16:54:50 -080024 - Transitivity
David Howells108b42b2006-03-31 16:00:29 +010025
26 (*) Explicit kernel barriers.
27
28 - Compiler barrier.
Jarek Poplawski81fc6322007-05-23 13:58:20 -070029 - CPU memory barriers.
David Howells108b42b2006-03-31 16:00:29 +010030 - MMIO write barrier.
31
32 (*) Implicit kernel memory barriers.
33
34 - Locking functions.
35 - Interrupt disabling functions.
David Howells50fa6102009-04-28 15:01:38 +010036 - Sleep and wake-up functions.
David Howells108b42b2006-03-31 16:00:29 +010037 - Miscellaneous functions.
38
39 (*) Inter-CPU locking barrier effects.
40
41 - Locks vs memory accesses.
42 - Locks vs I/O accesses.
43
44 (*) Where are memory barriers needed?
45
46 - Interprocessor interaction.
47 - Atomic operations.
48 - Accessing devices.
49 - Interrupts.
50
51 (*) Kernel I/O barrier effects.
52
53 (*) Assumed minimum execution ordering model.
54
55 (*) The effects of the cpu cache.
56
57 - Cache coherency.
58 - Cache coherency vs DMA.
59 - Cache coherency vs MMIO.
60
61 (*) The things CPUs get up to.
62
63 - And then there's the Alpha.
64
David Howells90fddab2010-03-24 09:43:00 +000065 (*) Example uses.
66
67 - Circular buffers.
68
David Howells108b42b2006-03-31 16:00:29 +010069 (*) References.
70
71
72============================
73ABSTRACT MEMORY ACCESS MODEL
74============================
75
76Consider the following abstract model of the system:
77
78 : :
79 : :
80 : :
81 +-------+ : +--------+ : +-------+
82 | | : | | : | |
83 | | : | | : | |
84 | CPU 1 |<----->| Memory |<----->| CPU 2 |
85 | | : | | : | |
86 | | : | | : | |
87 +-------+ : +--------+ : +-------+
88 ^ : ^ : ^
89 | : | : |
90 | : | : |
91 | : v : |
92 | : +--------+ : |
93 | : | | : |
94 | : | | : |
95 +---------->| Device |<----------+
96 : | | :
97 : | | :
98 : +--------+ :
99 : :
100
101Each CPU executes a program that generates memory access operations. In the
102abstract CPU, memory operation ordering is very relaxed, and a CPU may actually
103perform the memory operations in any order it likes, provided program causality
104appears to be maintained. Similarly, the compiler may also arrange the
105instructions it emits in any order it likes, provided it doesn't affect the
106apparent operation of the program.
107
108So in the above diagram, the effects of the memory operations performed by a
109CPU are perceived by the rest of the system as the operations cross the
110interface between the CPU and rest of the system (the dotted lines).
111
112
113For example, consider the following sequence of events:
114
115 CPU 1 CPU 2
116 =============== ===============
117 { A == 1; B == 2 }
Alexey Dobriyan615cc2c2014-06-06 14:36:41 -0700118 A = 3; x = B;
119 B = 4; y = A;
David Howells108b42b2006-03-31 16:00:29 +0100120
121The set of accesses as seen by the memory system in the middle can be arranged
122in 24 different combinations:
123
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400124 STORE A=3, STORE B=4, y=LOAD A->3, x=LOAD B->4
125 STORE A=3, STORE B=4, x=LOAD B->4, y=LOAD A->3
126 STORE A=3, y=LOAD A->3, STORE B=4, x=LOAD B->4
127 STORE A=3, y=LOAD A->3, x=LOAD B->2, STORE B=4
128 STORE A=3, x=LOAD B->2, STORE B=4, y=LOAD A->3
129 STORE A=3, x=LOAD B->2, y=LOAD A->3, STORE B=4
130 STORE B=4, STORE A=3, y=LOAD A->3, x=LOAD B->4
David Howells108b42b2006-03-31 16:00:29 +0100131 STORE B=4, ...
132 ...
133
134and can thus result in four different combinations of values:
135
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400136 x == 2, y == 1
137 x == 2, y == 3
138 x == 4, y == 1
139 x == 4, y == 3
David Howells108b42b2006-03-31 16:00:29 +0100140
141
142Furthermore, the stores committed by a CPU to the memory system may not be
143perceived by the loads made by another CPU in the same order as the stores were
144committed.
145
146
147As a further example, consider this sequence of events:
148
149 CPU 1 CPU 2
150 =============== ===============
151 { A == 1, B == 2, C = 3, P == &A, Q == &C }
152 B = 4; Q = P;
153 P = &B D = *Q;
154
155There is an obvious data dependency here, as the value loaded into D depends on
156the address retrieved from P by CPU 2. At the end of the sequence, any of the
157following results are possible:
158
159 (Q == &A) and (D == 1)
160 (Q == &B) and (D == 2)
161 (Q == &B) and (D == 4)
162
163Note that CPU 2 will never try and load C into D because the CPU will load P
164into Q before issuing the load of *Q.
165
166
167DEVICE OPERATIONS
168-----------------
169
170Some devices present their control interfaces as collections of memory
171locations, but the order in which the control registers are accessed is very
172important. For instance, imagine an ethernet card with a set of internal
173registers that are accessed through an address port register (A) and a data
174port register (D). To read internal register 5, the following code might then
175be used:
176
177 *A = 5;
178 x = *D;
179
180but this might show up as either of the following two sequences:
181
182 STORE *A = 5, x = LOAD *D
183 x = LOAD *D, STORE *A = 5
184
185the second of which will almost certainly result in a malfunction, since it set
186the address _after_ attempting to read the register.
187
188
189GUARANTEES
190----------
191
192There are some minimal guarantees that may be expected of a CPU:
193
194 (*) On any given CPU, dependent memory accesses will be issued in order, with
195 respect to itself. This means that for:
196
Chris Metcalff84cfbb2015-11-23 17:04:17 -0500197 Q = READ_ONCE(P); smp_read_barrier_depends(); D = READ_ONCE(*Q);
David Howells108b42b2006-03-31 16:00:29 +0100198
199 the CPU will issue the following memory operations:
200
201 Q = LOAD P, D = LOAD *Q
202
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800203 and always in that order. On most systems, smp_read_barrier_depends()
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700204 does nothing, but it is required for DEC Alpha. The READ_ONCE()
Chris Metcalff84cfbb2015-11-23 17:04:17 -0500205 is required to prevent compiler mischief. Please note that you
206 should normally use something like rcu_dereference() instead of
207 open-coding smp_read_barrier_depends().
David Howells108b42b2006-03-31 16:00:29 +0100208
209 (*) Overlapping loads and stores within a particular CPU will appear to be
210 ordered within that CPU. This means that for:
211
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700212 a = READ_ONCE(*X); WRITE_ONCE(*X, b);
David Howells108b42b2006-03-31 16:00:29 +0100213
214 the CPU will only issue the following sequence of memory operations:
215
216 a = LOAD *X, STORE *X = b
217
218 And for:
219
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700220 WRITE_ONCE(*X, c); d = READ_ONCE(*X);
David Howells108b42b2006-03-31 16:00:29 +0100221
222 the CPU will only issue:
223
224 STORE *X = c, d = LOAD *X
225
Matt LaPlantefa00e7e2006-11-30 04:55:36 +0100226 (Loads and stores overlap if they are targeted at overlapping pieces of
David Howells108b42b2006-03-31 16:00:29 +0100227 memory).
228
229And there are a number of things that _must_ or _must_not_ be assumed:
230
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700231 (*) It _must_not_ be assumed that the compiler will do what you want
232 with memory references that are not protected by READ_ONCE() and
233 WRITE_ONCE(). Without them, the compiler is within its rights to
234 do all sorts of "creative" transformations, which are covered in
Paul E. McKenney895f5542016-01-06 14:23:03 -0800235 the COMPILER BARRIER section.
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800236
David Howells108b42b2006-03-31 16:00:29 +0100237 (*) It _must_not_ be assumed that independent loads and stores will be issued
238 in the order given. This means that for:
239
240 X = *A; Y = *B; *D = Z;
241
242 we may get any of the following sequences:
243
244 X = LOAD *A, Y = LOAD *B, STORE *D = Z
245 X = LOAD *A, STORE *D = Z, Y = LOAD *B
246 Y = LOAD *B, X = LOAD *A, STORE *D = Z
247 Y = LOAD *B, STORE *D = Z, X = LOAD *A
248 STORE *D = Z, X = LOAD *A, Y = LOAD *B
249 STORE *D = Z, Y = LOAD *B, X = LOAD *A
250
251 (*) It _must_ be assumed that overlapping memory accesses may be merged or
252 discarded. This means that for:
253
254 X = *A; Y = *(A + 4);
255
256 we may get any one of the following sequences:
257
258 X = LOAD *A; Y = LOAD *(A + 4);
259 Y = LOAD *(A + 4); X = LOAD *A;
260 {X, Y} = LOAD {*A, *(A + 4) };
261
262 And for:
263
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700264 *A = X; *(A + 4) = Y;
David Howells108b42b2006-03-31 16:00:29 +0100265
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700266 we may get any of:
David Howells108b42b2006-03-31 16:00:29 +0100267
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700268 STORE *A = X; STORE *(A + 4) = Y;
269 STORE *(A + 4) = Y; STORE *A = X;
270 STORE {*A, *(A + 4) } = {X, Y};
David Howells108b42b2006-03-31 16:00:29 +0100271
Paul E. McKenney432fbf32014-09-04 17:12:49 -0700272And there are anti-guarantees:
273
274 (*) These guarantees do not apply to bitfields, because compilers often
275 generate code to modify these using non-atomic read-modify-write
276 sequences. Do not attempt to use bitfields to synchronize parallel
277 algorithms.
278
279 (*) Even in cases where bitfields are protected by locks, all fields
280 in a given bitfield must be protected by one lock. If two fields
281 in a given bitfield are protected by different locks, the compiler's
282 non-atomic read-modify-write sequences can cause an update to one
283 field to corrupt the value of an adjacent field.
284
285 (*) These guarantees apply only to properly aligned and sized scalar
286 variables. "Properly sized" currently means variables that are
287 the same size as "char", "short", "int" and "long". "Properly
288 aligned" means the natural alignment, thus no constraints for
289 "char", two-byte alignment for "short", four-byte alignment for
290 "int", and either four-byte or eight-byte alignment for "long",
291 on 32-bit and 64-bit systems, respectively. Note that these
292 guarantees were introduced into the C11 standard, so beware when
293 using older pre-C11 compilers (for example, gcc 4.6). The portion
294 of the standard containing this guarantee is Section 3.14, which
295 defines "memory location" as follows:
296
297 memory location
298 either an object of scalar type, or a maximal sequence
299 of adjacent bit-fields all having nonzero width
300
301 NOTE 1: Two threads of execution can update and access
302 separate memory locations without interfering with
303 each other.
304
305 NOTE 2: A bit-field and an adjacent non-bit-field member
306 are in separate memory locations. The same applies
307 to two bit-fields, if one is declared inside a nested
308 structure declaration and the other is not, or if the two
309 are separated by a zero-length bit-field declaration,
310 or if they are separated by a non-bit-field member
311 declaration. It is not safe to concurrently update two
312 bit-fields in the same structure if all members declared
313 between them are also bit-fields, no matter what the
314 sizes of those intervening bit-fields happen to be.
315
David Howells108b42b2006-03-31 16:00:29 +0100316
317=========================
318WHAT ARE MEMORY BARRIERS?
319=========================
320
321As can be seen above, independent memory operations are effectively performed
322in random order, but this can be a problem for CPU-CPU interaction and for I/O.
323What is required is some way of intervening to instruct the compiler and the
324CPU to restrict the order.
325
326Memory barriers are such interventions. They impose a perceived partial
David Howells2b948952006-06-25 05:48:49 -0700327ordering over the memory operations on either side of the barrier.
328
329Such enforcement is important because the CPUs and other devices in a system
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700330can use a variety of tricks to improve performance, including reordering,
David Howells2b948952006-06-25 05:48:49 -0700331deferral and combination of memory operations; speculative loads; speculative
332branch prediction and various types of caching. Memory barriers are used to
333override or suppress these tricks, allowing the code to sanely control the
334interaction of multiple CPUs and/or devices.
David Howells108b42b2006-03-31 16:00:29 +0100335
336
337VARIETIES OF MEMORY BARRIER
338---------------------------
339
340Memory barriers come in four basic varieties:
341
342 (1) Write (or store) memory barriers.
343
344 A write memory barrier gives a guarantee that all the STORE operations
345 specified before the barrier will appear to happen before all the STORE
346 operations specified after the barrier with respect to the other
347 components of the system.
348
349 A write barrier is a partial ordering on stores only; it is not required
350 to have any effect on loads.
351
David Howells6bc39272006-06-25 05:49:22 -0700352 A CPU can be viewed as committing a sequence of store operations to the
David Howells108b42b2006-03-31 16:00:29 +0100353 memory system as time progresses. All stores before a write barrier will
354 occur in the sequence _before_ all the stores after the write barrier.
355
356 [!] Note that write barriers should normally be paired with read or data
357 dependency barriers; see the "SMP barrier pairing" subsection.
358
359
360 (2) Data dependency barriers.
361
362 A data dependency barrier is a weaker form of read barrier. In the case
363 where two loads are performed such that the second depends on the result
364 of the first (eg: the first load retrieves the address to which the second
365 load will be directed), a data dependency barrier would be required to
366 make sure that the target of the second load is updated before the address
367 obtained by the first load is accessed.
368
369 A data dependency barrier is a partial ordering on interdependent loads
370 only; it is not required to have any effect on stores, independent loads
371 or overlapping loads.
372
373 As mentioned in (1), the other CPUs in the system can be viewed as
374 committing sequences of stores to the memory system that the CPU being
375 considered can then perceive. A data dependency barrier issued by the CPU
376 under consideration guarantees that for any load preceding it, if that
377 load touches one of a sequence of stores from another CPU, then by the
378 time the barrier completes, the effects of all the stores prior to that
379 touched by the load will be perceptible to any loads issued after the data
380 dependency barrier.
381
382 See the "Examples of memory barrier sequences" subsection for diagrams
383 showing the ordering constraints.
384
385 [!] Note that the first load really has to have a _data_ dependency and
386 not a control dependency. If the address for the second load is dependent
387 on the first load, but the dependency is through a conditional rather than
388 actually loading the address itself, then it's a _control_ dependency and
389 a full read barrier or better is required. See the "Control dependencies"
390 subsection for more information.
391
392 [!] Note that data dependency barriers should normally be paired with
393 write barriers; see the "SMP barrier pairing" subsection.
394
395
396 (3) Read (or load) memory barriers.
397
398 A read barrier is a data dependency barrier plus a guarantee that all the
399 LOAD operations specified before the barrier will appear to happen before
400 all the LOAD operations specified after the barrier with respect to the
401 other components of the system.
402
403 A read barrier is a partial ordering on loads only; it is not required to
404 have any effect on stores.
405
406 Read memory barriers imply data dependency barriers, and so can substitute
407 for them.
408
409 [!] Note that read barriers should normally be paired with write barriers;
410 see the "SMP barrier pairing" subsection.
411
412
413 (4) General memory barriers.
414
David Howells670bd952006-06-10 09:54:12 -0700415 A general memory barrier gives a guarantee that all the LOAD and STORE
416 operations specified before the barrier will appear to happen before all
417 the LOAD and STORE operations specified after the barrier with respect to
418 the other components of the system.
419
420 A general memory barrier is a partial ordering over both loads and stores.
David Howells108b42b2006-03-31 16:00:29 +0100421
422 General memory barriers imply both read and write memory barriers, and so
423 can substitute for either.
424
425
426And a couple of implicit varieties:
427
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100428 (5) ACQUIRE operations.
David Howells108b42b2006-03-31 16:00:29 +0100429
430 This acts as a one-way permeable barrier. It guarantees that all memory
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100431 operations after the ACQUIRE operation will appear to happen after the
432 ACQUIRE operation with respect to the other components of the system.
433 ACQUIRE operations include LOCK operations and smp_load_acquire()
434 operations.
David Howells108b42b2006-03-31 16:00:29 +0100435
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100436 Memory operations that occur before an ACQUIRE operation may appear to
437 happen after it completes.
David Howells108b42b2006-03-31 16:00:29 +0100438
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100439 An ACQUIRE operation should almost always be paired with a RELEASE
440 operation.
David Howells108b42b2006-03-31 16:00:29 +0100441
442
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100443 (6) RELEASE operations.
David Howells108b42b2006-03-31 16:00:29 +0100444
445 This also acts as a one-way permeable barrier. It guarantees that all
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100446 memory operations before the RELEASE operation will appear to happen
447 before the RELEASE operation with respect to the other components of the
448 system. RELEASE operations include UNLOCK operations and
449 smp_store_release() operations.
David Howells108b42b2006-03-31 16:00:29 +0100450
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100451 Memory operations that occur after a RELEASE operation may appear to
David Howells108b42b2006-03-31 16:00:29 +0100452 happen before it completes.
453
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100454 The use of ACQUIRE and RELEASE operations generally precludes the need
455 for other sorts of memory barrier (but note the exceptions mentioned in
456 the subsection "MMIO write barrier"). In addition, a RELEASE+ACQUIRE
457 pair is -not- guaranteed to act as a full memory barrier. However, after
458 an ACQUIRE on a given variable, all memory accesses preceding any prior
459 RELEASE on that same variable are guaranteed to be visible. In other
460 words, within a given variable's critical section, all accesses of all
461 previous critical sections for that variable are guaranteed to have
462 completed.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -0800463
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100464 This means that ACQUIRE acts as a minimal "acquire" operation and
465 RELEASE acts as a minimal "release" operation.
David Howells108b42b2006-03-31 16:00:29 +0100466
467
468Memory barriers are only required where there's a possibility of interaction
469between two CPUs or between a CPU and a device. If it can be guaranteed that
470there won't be any such interaction in any particular piece of code, then
471memory barriers are unnecessary in that piece of code.
472
473
474Note that these are the _minimum_ guarantees. Different architectures may give
475more substantial guarantees, but they may _not_ be relied upon outside of arch
476specific code.
477
478
479WHAT MAY NOT BE ASSUMED ABOUT MEMORY BARRIERS?
480----------------------------------------------
481
482There are certain things that the Linux kernel memory barriers do not guarantee:
483
484 (*) There is no guarantee that any of the memory accesses specified before a
485 memory barrier will be _complete_ by the completion of a memory barrier
486 instruction; the barrier can be considered to draw a line in that CPU's
487 access queue that accesses of the appropriate type may not cross.
488
489 (*) There is no guarantee that issuing a memory barrier on one CPU will have
490 any direct effect on another CPU or any other hardware in the system. The
491 indirect effect will be the order in which the second CPU sees the effects
492 of the first CPU's accesses occur, but see the next point:
493
David Howells6bc39272006-06-25 05:49:22 -0700494 (*) There is no guarantee that a CPU will see the correct order of effects
David Howells108b42b2006-03-31 16:00:29 +0100495 from a second CPU's accesses, even _if_ the second CPU uses a memory
496 barrier, unless the first CPU _also_ uses a matching memory barrier (see
497 the subsection on "SMP Barrier Pairing").
498
499 (*) There is no guarantee that some intervening piece of off-the-CPU
500 hardware[*] will not reorder the memory accesses. CPU cache coherency
501 mechanisms should propagate the indirect effects of a memory barrier
502 between CPUs, but might not do so in order.
503
504 [*] For information on bus mastering DMA and coherency please read:
505
Randy Dunlap4b5ff462008-03-10 17:16:32 -0700506 Documentation/PCI/pci.txt
Paul Bolle395cf962011-08-15 02:02:26 +0200507 Documentation/DMA-API-HOWTO.txt
David Howells108b42b2006-03-31 16:00:29 +0100508 Documentation/DMA-API.txt
509
510
511DATA DEPENDENCY BARRIERS
512------------------------
513
514The usage requirements of data dependency barriers are a little subtle, and
515it's not always obvious that they're needed. To illustrate, consider the
516following sequence of events:
517
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800518 CPU 1 CPU 2
519 =============== ===============
David Howells108b42b2006-03-31 16:00:29 +0100520 { A == 1, B == 2, C = 3, P == &A, Q == &C }
521 B = 4;
522 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700523 WRITE_ONCE(P, &B)
524 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800525 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100526
527There's a clear data dependency here, and it would seem that by the end of the
528sequence, Q must be either &A or &B, and that:
529
530 (Q == &A) implies (D == 1)
531 (Q == &B) implies (D == 4)
532
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700533But! CPU 2's perception of P may be updated _before_ its perception of B, thus
David Howells108b42b2006-03-31 16:00:29 +0100534leading to the following situation:
535
536 (Q == &B) and (D == 2) ????
537
538Whilst this may seem like a failure of coherency or causality maintenance, it
539isn't, and this behaviour can be observed on certain real CPUs (such as the DEC
540Alpha).
541
David Howells2b948952006-06-25 05:48:49 -0700542To deal with this, a data dependency barrier or better must be inserted
543between the address load and the data load:
David Howells108b42b2006-03-31 16:00:29 +0100544
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800545 CPU 1 CPU 2
546 =============== ===============
David Howells108b42b2006-03-31 16:00:29 +0100547 { A == 1, B == 2, C = 3, P == &A, Q == &C }
548 B = 4;
549 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700550 WRITE_ONCE(P, &B);
551 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800552 <data dependency barrier>
553 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100554
555This enforces the occurrence of one of the two implications, and prevents the
556third possibility from arising.
557
Paul E. McKenney92a84dd2016-01-14 14:17:04 -0800558A data-dependency barrier must also order against dependent writes:
559
560 CPU 1 CPU 2
561 =============== ===============
562 { A == 1, B == 2, C = 3, P == &A, Q == &C }
563 B = 4;
564 <write barrier>
565 WRITE_ONCE(P, &B);
566 Q = READ_ONCE(P);
567 <data dependency barrier>
568 *Q = 5;
569
570The data-dependency barrier must order the read into Q with the store
571into *Q. This prohibits this outcome:
572
573 (Q == B) && (B == 4)
574
575Please note that this pattern should be rare. After all, the whole point
576of dependency ordering is to -prevent- writes to the data structure, along
577with the expensive cache misses associated with those writes. This pattern
578can be used to record rare error conditions and the like, and the ordering
579prevents such records from being lost.
580
581
David Howells108b42b2006-03-31 16:00:29 +0100582[!] Note that this extremely counterintuitive situation arises most easily on
583machines with split caches, so that, for example, one cache bank processes
584even-numbered cache lines and the other bank processes odd-numbered cache
585lines. The pointer P might be stored in an odd-numbered cache line, and the
586variable B might be stored in an even-numbered cache line. Then, if the
587even-numbered bank of the reading CPU's cache is extremely busy while the
588odd-numbered bank is idle, one can see the new value of the pointer P (&B),
David Howells6bc39272006-06-25 05:49:22 -0700589but the old value of the variable B (2).
David Howells108b42b2006-03-31 16:00:29 +0100590
591
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800592The data dependency barrier is very important to the RCU system,
593for example. See rcu_assign_pointer() and rcu_dereference() in
594include/linux/rcupdate.h. This permits the current target of an RCU'd
595pointer to be replaced with a new modified target, without the replacement
596target appearing to be incompletely initialised.
David Howells108b42b2006-03-31 16:00:29 +0100597
598See also the subsection on "Cache Coherency" for a more thorough example.
599
600
601CONTROL DEPENDENCIES
602--------------------
603
Paul E. McKenneyff382812015-02-17 10:00:06 -0800604A load-load control dependency requires a full read memory barrier, not
605simply a data dependency barrier to make it work correctly. Consider the
606following bit of code:
David Howells108b42b2006-03-31 16:00:29 +0100607
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700608 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800609 if (q) {
610 <data dependency barrier> /* BUG: No data dependency!!! */
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700611 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700612 }
David Howells108b42b2006-03-31 16:00:29 +0100613
614This will not have the desired effect because there is no actual data
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800615dependency, but rather a control dependency that the CPU may short-circuit
616by attempting to predict the outcome in advance, so that other CPUs see
617the load from b as having happened before the load from a. In such a
618case what's actually required is:
David Howells108b42b2006-03-31 16:00:29 +0100619
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700620 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800621 if (q) {
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700622 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700623 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700624 }
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800625
626However, stores are not speculated. This means that ordering -is- provided
Paul E. McKenneyff382812015-02-17 10:00:06 -0800627for load-store control dependencies, as in the following example:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800628
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800629 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700630 if (q) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700631 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800632 }
633
Paul E. McKenney5af46922015-04-25 12:48:29 -0700634Control dependencies pair normally with other types of barriers. That
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800635said, please note that READ_ONCE() is not optional! Without the
636READ_ONCE(), the compiler might combine the load from 'a' with other
637loads from 'a', and the store to 'b' with other stores to 'b', with
638possible highly counterintuitive effects on ordering.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800639
640Worse yet, if the compiler is able to prove (say) that the value of
641variable 'a' is always non-zero, it would be well within its rights
642to optimize the original example by eliminating the "if" statement
643as follows:
644
645 q = a;
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700646 b = p; /* BUG: Compiler and CPU can both reorder!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800647
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800648So don't leave out the READ_ONCE().
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700649
650It is tempting to try to enforce ordering on identical stores on both
651branches of the "if" statement as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800652
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800653 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800654 if (q) {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800655 barrier();
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700656 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800657 do_something();
658 } else {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800659 barrier();
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700660 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800661 do_something_else();
662 }
663
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700664Unfortunately, current compilers will transform this as follows at high
665optimization levels:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800666
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800667 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700668 barrier();
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700669 WRITE_ONCE(b, p); /* BUG: No ordering vs. load from a!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800670 if (q) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700671 /* WRITE_ONCE(b, p); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800672 do_something();
673 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700674 /* WRITE_ONCE(b, p); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800675 do_something_else();
676 }
677
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700678Now there is no conditional between the load from 'a' and the store to
679'b', which means that the CPU is within its rights to reorder them:
680The conditional is absolutely required, and must be present in the
681assembly code even after all compiler optimizations have been applied.
682Therefore, if you need ordering in this example, you need explicit
683memory barriers, for example, smp_store_release():
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800684
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700685 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700686 if (q) {
687 smp_store_release(&b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800688 do_something();
689 } else {
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700690 smp_store_release(&b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800691 do_something_else();
692 }
693
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700694In contrast, without explicit memory barriers, two-legged-if control
695ordering is guaranteed only when the stores differ, for example:
696
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800697 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700698 if (q) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700699 WRITE_ONCE(b, p);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700700 do_something();
701 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700702 WRITE_ONCE(b, r);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700703 do_something_else();
704 }
705
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800706The initial READ_ONCE() is still required to prevent the compiler from
707proving the value of 'a'.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800708
709In addition, you need to be careful what you do with the local variable 'q',
710otherwise the compiler might be able to guess the value and again remove
711the needed conditional. For example:
712
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800713 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800714 if (q % MAX) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700715 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800716 do_something();
717 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700718 WRITE_ONCE(b, r);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800719 do_something_else();
720 }
721
722If MAX is defined to be 1, then the compiler knows that (q % MAX) is
723equal to zero, in which case the compiler is within its rights to
724transform the above code into the following:
725
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800726 q = READ_ONCE(a);
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700727 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800728 do_something_else();
729
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700730Given this transformation, the CPU is not required to respect the ordering
731between the load from variable 'a' and the store to variable 'b'. It is
732tempting to add a barrier(), but this does not help. The conditional
733is gone, and the barrier won't bring it back. Therefore, if you are
734relying on this ordering, you should make sure that MAX is greater than
735one, perhaps as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800736
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800737 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800738 BUILD_BUG_ON(MAX <= 1); /* Order load from a with store to b. */
739 if (q % MAX) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700740 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800741 do_something();
742 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700743 WRITE_ONCE(b, r);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800744 do_something_else();
745 }
746
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700747Please note once again that the stores to 'b' differ. If they were
748identical, as noted earlier, the compiler could pull this store outside
749of the 'if' statement.
750
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700751You must also be careful not to rely too much on boolean short-circuit
752evaluation. Consider this example:
753
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800754 q = READ_ONCE(a);
Paul E. McKenney57aecae2015-05-18 18:27:42 -0700755 if (q || 1 > 0)
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700756 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700757
Paul E. McKenney5af46922015-04-25 12:48:29 -0700758Because the first condition cannot fault and the second condition is
759always true, the compiler can transform this example as following,
760defeating control dependency:
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700761
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800762 q = READ_ONCE(a);
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700763 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700764
765This example underscores the need to ensure that the compiler cannot
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700766out-guess your code. More generally, although READ_ONCE() does force
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700767the compiler to actually emit code for a given load, it does not force
768the compiler to use the results.
769
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800770Finally, control dependencies do -not- provide transitivity. This is
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700771demonstrated by two related examples, with the initial values of
772x and y both being zero:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800773
774 CPU 0 CPU 1
Paul E. McKenney5af46922015-04-25 12:48:29 -0700775 ======================= =======================
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800776 r1 = READ_ONCE(x); r2 = READ_ONCE(y);
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700777 if (r1 > 0) if (r2 > 0)
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700778 WRITE_ONCE(y, 1); WRITE_ONCE(x, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800779
780 assert(!(r1 == 1 && r2 == 1));
781
782The above two-CPU example will never trigger the assert(). However,
783if control dependencies guaranteed transitivity (which they do not),
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700784then adding the following CPU would guarantee a related assertion:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800785
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700786 CPU 2
787 =====================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700788 WRITE_ONCE(x, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800789
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700790 assert(!(r1 == 2 && r2 == 1 && x == 2)); /* FAILS!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800791
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700792But because control dependencies do -not- provide transitivity, the above
793assertion can fail after the combined three-CPU example completes. If you
794need the three-CPU example to provide ordering, you will need smp_mb()
795between the loads and stores in the CPU 0 and CPU 1 code fragments,
Paul E. McKenney5af46922015-04-25 12:48:29 -0700796that is, just before or just after the "if" statements. Furthermore,
797the original two-CPU example is very fragile and should be avoided.
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700798
799These two examples are the LB and WWC litmus tests from this paper:
800http://www.cl.cam.ac.uk/users/pes20/ppc-supplemental/test6.pdf and this
801site: https://www.cl.cam.ac.uk/~pes20/ppcmem/index.html.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800802
803In summary:
804
805 (*) Control dependencies can order prior loads against later stores.
806 However, they do -not- guarantee any other sort of ordering:
807 Not prior loads against later loads, nor prior stores against
808 later anything. If you need these other forms of ordering,
Davidlohr Buesod87510c2014-12-28 01:11:16 -0800809 use smp_rmb(), smp_wmb(), or, in the case of prior stores and
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800810 later loads, smp_mb().
811
Paul E. McKenney7817b792015-12-29 16:23:18 -0800812 (*) If both legs of the "if" statement begin with identical stores to
813 the same variable, then those stores must be ordered, either by
814 preceding both of them with smp_mb() or by using smp_store_release()
815 to carry out the stores. Please note that it is -not- sufficient
Paul E. McKenneya5052652016-04-12 08:52:49 -0700816 to use barrier() at beginning of each leg of the "if" statement
817 because, as shown by the example above, optimizing compilers can
818 destroy the control dependency while respecting the letter of the
819 barrier() law.
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800820
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800821 (*) Control dependencies require at least one run-time conditional
Paul E. McKenney586dd562014-02-11 12:28:06 -0800822 between the prior load and the subsequent store, and this
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700823 conditional must involve the prior load. If the compiler is able
824 to optimize the conditional away, it will have also optimized
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800825 away the ordering. Careful use of READ_ONCE() and WRITE_ONCE()
826 can help to preserve the needed conditional.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800827
828 (*) Control dependencies require that the compiler avoid reordering the
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800829 dependency into nonexistence. Careful use of READ_ONCE() or
830 atomic{,64}_read() can help to preserve your control dependency.
Paul E. McKenney895f5542016-01-06 14:23:03 -0800831 Please see the COMPILER BARRIER section for more information.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800832
Paul E. McKenneyff382812015-02-17 10:00:06 -0800833 (*) Control dependencies pair normally with other types of barriers.
834
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800835 (*) Control dependencies do -not- provide transitivity. If you
836 need transitivity, use smp_mb().
David Howells108b42b2006-03-31 16:00:29 +0100837
838
839SMP BARRIER PAIRING
840-------------------
841
842When dealing with CPU-CPU interactions, certain types of memory barrier should
843always be paired. A lack of appropriate pairing is almost certainly an error.
844
Paul E. McKenneyff382812015-02-17 10:00:06 -0800845General barriers pair with each other, though they also pair with most
846other types of barriers, albeit without transitivity. An acquire barrier
847pairs with a release barrier, but both may also pair with other barriers,
848including of course general barriers. A write barrier pairs with a data
849dependency barrier, a control dependency, an acquire barrier, a release
850barrier, a read barrier, or a general barrier. Similarly a read barrier,
851control dependency, or a data dependency barrier pairs with a write
852barrier, an acquire barrier, a release barrier, or a general barrier:
David Howells108b42b2006-03-31 16:00:29 +0100853
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800854 CPU 1 CPU 2
855 =============== ===============
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700856 WRITE_ONCE(a, 1);
David Howells108b42b2006-03-31 16:00:29 +0100857 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700858 WRITE_ONCE(b, 2); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800859 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700860 y = READ_ONCE(a);
David Howells108b42b2006-03-31 16:00:29 +0100861
862Or:
863
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800864 CPU 1 CPU 2
865 =============== ===============================
David Howells108b42b2006-03-31 16:00:29 +0100866 a = 1;
867 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700868 WRITE_ONCE(b, &a); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800869 <data dependency barrier>
870 y = *x;
David Howells108b42b2006-03-31 16:00:29 +0100871
Paul E. McKenneyff382812015-02-17 10:00:06 -0800872Or even:
873
874 CPU 1 CPU 2
875 =============== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700876 r1 = READ_ONCE(y);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800877 <general barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700878 WRITE_ONCE(y, 1); if (r2 = READ_ONCE(x)) {
Paul E. McKenneyff382812015-02-17 10:00:06 -0800879 <implicit control dependency>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700880 WRITE_ONCE(y, 1);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800881 }
882
883 assert(r1 == 0 || r2 == 0);
884
David Howells108b42b2006-03-31 16:00:29 +0100885Basically, the read barrier always has to be there, even though it can be of
886the "weaker" type.
887
David Howells670bd952006-06-10 09:54:12 -0700888[!] Note that the stores before the write barrier would normally be expected to
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700889match the loads after the read barrier or the data dependency barrier, and vice
David Howells670bd952006-06-10 09:54:12 -0700890versa:
891
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800892 CPU 1 CPU 2
893 =================== ===================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700894 WRITE_ONCE(a, 1); }---- --->{ v = READ_ONCE(c);
895 WRITE_ONCE(b, 2); } \ / { w = READ_ONCE(d);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800896 <write barrier> \ <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700897 WRITE_ONCE(c, 3); } / \ { x = READ_ONCE(a);
898 WRITE_ONCE(d, 4); }---- --->{ y = READ_ONCE(b);
David Howells670bd952006-06-10 09:54:12 -0700899
David Howells108b42b2006-03-31 16:00:29 +0100900
901EXAMPLES OF MEMORY BARRIER SEQUENCES
902------------------------------------
903
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700904Firstly, write barriers act as partial orderings on store operations.
David Howells108b42b2006-03-31 16:00:29 +0100905Consider the following sequence of events:
906
907 CPU 1
908 =======================
909 STORE A = 1
910 STORE B = 2
911 STORE C = 3
912 <write barrier>
913 STORE D = 4
914 STORE E = 5
915
916This sequence of events is committed to the memory coherence system in an order
917that the rest of the system might perceive as the unordered set of { STORE A,
Adrian Bunk80f72282006-06-30 18:27:16 +0200918STORE B, STORE C } all occurring before the unordered set of { STORE D, STORE E
David Howells108b42b2006-03-31 16:00:29 +0100919}:
920
921 +-------+ : :
922 | | +------+
923 | |------>| C=3 | } /\
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700924 | | : +------+ }----- \ -----> Events perceptible to
925 | | : | A=1 | } \/ the rest of the system
David Howells108b42b2006-03-31 16:00:29 +0100926 | | : +------+ }
927 | CPU 1 | : | B=2 | }
928 | | +------+ }
929 | | wwwwwwwwwwwwwwww } <--- At this point the write barrier
930 | | +------+ } requires all stores prior to the
931 | | : | E=5 | } barrier to be committed before
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700932 | | : +------+ } further stores may take place
David Howells108b42b2006-03-31 16:00:29 +0100933 | |------>| D=4 | }
934 | | +------+
935 +-------+ : :
936 |
David Howells670bd952006-06-10 09:54:12 -0700937 | Sequence in which stores are committed to the
938 | memory system by CPU 1
David Howells108b42b2006-03-31 16:00:29 +0100939 V
940
941
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700942Secondly, data dependency barriers act as partial orderings on data-dependent
David Howells108b42b2006-03-31 16:00:29 +0100943loads. Consider the following sequence of events:
944
945 CPU 1 CPU 2
946 ======================= =======================
David Howellsc14038c2006-04-10 22:54:24 -0700947 { B = 7; X = 9; Y = 8; C = &Y }
David Howells108b42b2006-03-31 16:00:29 +0100948 STORE A = 1
949 STORE B = 2
950 <write barrier>
951 STORE C = &B LOAD X
952 STORE D = 4 LOAD C (gets &B)
953 LOAD *C (reads B)
954
955Without intervention, CPU 2 may perceive the events on CPU 1 in some
956effectively random order, despite the write barrier issued by CPU 1:
957
958 +-------+ : : : :
959 | | +------+ +-------+ | Sequence of update
960 | |------>| B=2 |----- --->| Y->8 | | of perception on
961 | | : +------+ \ +-------+ | CPU 2
962 | CPU 1 | : | A=1 | \ --->| C->&Y | V
963 | | +------+ | +-------+
964 | | wwwwwwwwwwwwwwww | : :
965 | | +------+ | : :
966 | | : | C=&B |--- | : : +-------+
967 | | : +------+ \ | +-------+ | |
968 | |------>| D=4 | ----------->| C->&B |------>| |
969 | | +------+ | +-------+ | |
970 +-------+ : : | : : | |
971 | : : | |
972 | : : | CPU 2 |
973 | +-------+ | |
974 Apparently incorrect ---> | | B->7 |------>| |
975 perception of B (!) | +-------+ | |
976 | : : | |
977 | +-------+ | |
978 The load of X holds ---> \ | X->9 |------>| |
979 up the maintenance \ +-------+ | |
980 of coherence of B ----->| B->2 | +-------+
981 +-------+
982 : :
983
984
985In the above example, CPU 2 perceives that B is 7, despite the load of *C
Paolo Ornati670e9f32006-10-03 22:57:56 +0200986(which would be B) coming after the LOAD of C.
David Howells108b42b2006-03-31 16:00:29 +0100987
988If, however, a data dependency barrier were to be placed between the load of C
David Howellsc14038c2006-04-10 22:54:24 -0700989and the load of *C (ie: B) on CPU 2:
990
991 CPU 1 CPU 2
992 ======================= =======================
993 { B = 7; X = 9; Y = 8; C = &Y }
994 STORE A = 1
995 STORE B = 2
996 <write barrier>
997 STORE C = &B LOAD X
998 STORE D = 4 LOAD C (gets &B)
999 <data dependency barrier>
1000 LOAD *C (reads B)
1001
1002then the following will occur:
David Howells108b42b2006-03-31 16:00:29 +01001003
1004 +-------+ : : : :
1005 | | +------+ +-------+
1006 | |------>| B=2 |----- --->| Y->8 |
1007 | | : +------+ \ +-------+
1008 | CPU 1 | : | A=1 | \ --->| C->&Y |
1009 | | +------+ | +-------+
1010 | | wwwwwwwwwwwwwwww | : :
1011 | | +------+ | : :
1012 | | : | C=&B |--- | : : +-------+
1013 | | : +------+ \ | +-------+ | |
1014 | |------>| D=4 | ----------->| C->&B |------>| |
1015 | | +------+ | +-------+ | |
1016 +-------+ : : | : : | |
1017 | : : | |
1018 | : : | CPU 2 |
1019 | +-------+ | |
David Howells670bd952006-06-10 09:54:12 -07001020 | | X->9 |------>| |
1021 | +-------+ | |
1022 Makes sure all effects ---> \ ddddddddddddddddd | |
1023 prior to the store of C \ +-------+ | |
1024 are perceptible to ----->| B->2 |------>| |
1025 subsequent loads +-------+ | |
David Howells108b42b2006-03-31 16:00:29 +01001026 : : +-------+
1027
1028
1029And thirdly, a read barrier acts as a partial order on loads. Consider the
1030following sequence of events:
1031
1032 CPU 1 CPU 2
1033 ======================= =======================
David Howells670bd952006-06-10 09:54:12 -07001034 { A = 0, B = 9 }
David Howells108b42b2006-03-31 16:00:29 +01001035 STORE A=1
David Howells108b42b2006-03-31 16:00:29 +01001036 <write barrier>
David Howells670bd952006-06-10 09:54:12 -07001037 STORE B=2
David Howells108b42b2006-03-31 16:00:29 +01001038 LOAD B
David Howells670bd952006-06-10 09:54:12 -07001039 LOAD A
David Howells108b42b2006-03-31 16:00:29 +01001040
1041Without intervention, CPU 2 may then choose to perceive the events on CPU 1 in
1042some effectively random order, despite the write barrier issued by CPU 1:
1043
David Howells670bd952006-06-10 09:54:12 -07001044 +-------+ : : : :
1045 | | +------+ +-------+
1046 | |------>| A=1 |------ --->| A->0 |
1047 | | +------+ \ +-------+
1048 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1049 | | +------+ | +-------+
1050 | |------>| B=2 |--- | : :
1051 | | +------+ \ | : : +-------+
1052 +-------+ : : \ | +-------+ | |
1053 ---------->| B->2 |------>| |
1054 | +-------+ | CPU 2 |
1055 | | A->0 |------>| |
1056 | +-------+ | |
1057 | : : +-------+
1058 \ : :
1059 \ +-------+
1060 ---->| A->1 |
1061 +-------+
1062 : :
David Howells108b42b2006-03-31 16:00:29 +01001063
1064
David Howells6bc39272006-06-25 05:49:22 -07001065If, however, a read barrier were to be placed between the load of B and the
David Howells670bd952006-06-10 09:54:12 -07001066load of A on CPU 2:
David Howells108b42b2006-03-31 16:00:29 +01001067
David Howells670bd952006-06-10 09:54:12 -07001068 CPU 1 CPU 2
1069 ======================= =======================
1070 { A = 0, B = 9 }
1071 STORE A=1
1072 <write barrier>
1073 STORE B=2
1074 LOAD B
1075 <read barrier>
1076 LOAD A
1077
1078then the partial ordering imposed by CPU 1 will be perceived correctly by CPU
10792:
1080
1081 +-------+ : : : :
1082 | | +------+ +-------+
1083 | |------>| A=1 |------ --->| A->0 |
1084 | | +------+ \ +-------+
1085 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1086 | | +------+ | +-------+
1087 | |------>| B=2 |--- | : :
1088 | | +------+ \ | : : +-------+
1089 +-------+ : : \ | +-------+ | |
1090 ---------->| B->2 |------>| |
1091 | +-------+ | CPU 2 |
1092 | : : | |
1093 | : : | |
1094 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1095 barrier causes all effects \ +-------+ | |
1096 prior to the storage of B ---->| A->1 |------>| |
1097 to be perceptible to CPU 2 +-------+ | |
1098 : : +-------+
1099
1100
1101To illustrate this more completely, consider what could happen if the code
1102contained a load of A either side of the read barrier:
1103
1104 CPU 1 CPU 2
1105 ======================= =======================
1106 { A = 0, B = 9 }
1107 STORE A=1
1108 <write barrier>
1109 STORE B=2
1110 LOAD B
1111 LOAD A [first load of A]
1112 <read barrier>
1113 LOAD A [second load of A]
1114
1115Even though the two loads of A both occur after the load of B, they may both
1116come up with different values:
1117
1118 +-------+ : : : :
1119 | | +------+ +-------+
1120 | |------>| A=1 |------ --->| A->0 |
1121 | | +------+ \ +-------+
1122 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1123 | | +------+ | +-------+
1124 | |------>| B=2 |--- | : :
1125 | | +------+ \ | : : +-------+
1126 +-------+ : : \ | +-------+ | |
1127 ---------->| B->2 |------>| |
1128 | +-------+ | CPU 2 |
1129 | : : | |
1130 | : : | |
1131 | +-------+ | |
1132 | | A->0 |------>| 1st |
1133 | +-------+ | |
1134 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1135 barrier causes all effects \ +-------+ | |
1136 prior to the storage of B ---->| A->1 |------>| 2nd |
1137 to be perceptible to CPU 2 +-------+ | |
1138 : : +-------+
1139
1140
1141But it may be that the update to A from CPU 1 becomes perceptible to CPU 2
1142before the read barrier completes anyway:
1143
1144 +-------+ : : : :
1145 | | +------+ +-------+
1146 | |------>| A=1 |------ --->| A->0 |
1147 | | +------+ \ +-------+
1148 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1149 | | +------+ | +-------+
1150 | |------>| B=2 |--- | : :
1151 | | +------+ \ | : : +-------+
1152 +-------+ : : \ | +-------+ | |
1153 ---------->| B->2 |------>| |
1154 | +-------+ | CPU 2 |
1155 | : : | |
1156 \ : : | |
1157 \ +-------+ | |
1158 ---->| A->1 |------>| 1st |
1159 +-------+ | |
1160 rrrrrrrrrrrrrrrrr | |
1161 +-------+ | |
1162 | A->1 |------>| 2nd |
1163 +-------+ | |
1164 : : +-------+
1165
1166
1167The guarantee is that the second load will always come up with A == 1 if the
1168load of B came up with B == 2. No such guarantee exists for the first load of
1169A; that may come up with either A == 0 or A == 1.
1170
1171
1172READ MEMORY BARRIERS VS LOAD SPECULATION
1173----------------------------------------
1174
1175Many CPUs speculate with loads: that is they see that they will need to load an
1176item from memory, and they find a time where they're not using the bus for any
1177other loads, and so do the load in advance - even though they haven't actually
1178got to that point in the instruction execution flow yet. This permits the
1179actual load instruction to potentially complete immediately because the CPU
1180already has the value to hand.
1181
1182It may turn out that the CPU didn't actually need the value - perhaps because a
1183branch circumvented the load - in which case it can discard the value or just
1184cache it for later use.
1185
1186Consider:
1187
Ingo Molnare0edc782013-11-22 11:24:53 +01001188 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001189 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001190 LOAD B
1191 DIVIDE } Divide instructions generally
1192 DIVIDE } take a long time to perform
1193 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001194
1195Which might appear as this:
1196
1197 : : +-------+
1198 +-------+ | |
1199 --->| B->2 |------>| |
1200 +-------+ | CPU 2 |
1201 : :DIVIDE | |
1202 +-------+ | |
1203 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1204 division speculates on the +-------+ ~ | |
1205 LOAD of A : : ~ | |
1206 : :DIVIDE | |
1207 : : ~ | |
1208 Once the divisions are complete --> : : ~-->| |
1209 the CPU can then perform the : : | |
1210 LOAD with immediate effect : : +-------+
1211
1212
1213Placing a read barrier or a data dependency barrier just before the second
1214load:
1215
Ingo Molnare0edc782013-11-22 11:24:53 +01001216 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001217 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001218 LOAD B
1219 DIVIDE
1220 DIVIDE
David Howells670bd952006-06-10 09:54:12 -07001221 <read barrier>
Ingo Molnare0edc782013-11-22 11:24:53 +01001222 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001223
1224will force any value speculatively obtained to be reconsidered to an extent
1225dependent on the type of barrier used. If there was no change made to the
1226speculated memory location, then the speculated value will just be used:
1227
1228 : : +-------+
1229 +-------+ | |
1230 --->| B->2 |------>| |
1231 +-------+ | CPU 2 |
1232 : :DIVIDE | |
1233 +-------+ | |
1234 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1235 division speculates on the +-------+ ~ | |
1236 LOAD of A : : ~ | |
1237 : :DIVIDE | |
1238 : : ~ | |
1239 : : ~ | |
1240 rrrrrrrrrrrrrrrr~ | |
1241 : : ~ | |
1242 : : ~-->| |
1243 : : | |
1244 : : +-------+
1245
1246
1247but if there was an update or an invalidation from another CPU pending, then
1248the speculation will be cancelled and the value reloaded:
1249
1250 : : +-------+
1251 +-------+ | |
1252 --->| B->2 |------>| |
1253 +-------+ | CPU 2 |
1254 : :DIVIDE | |
1255 +-------+ | |
1256 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1257 division speculates on the +-------+ ~ | |
1258 LOAD of A : : ~ | |
1259 : :DIVIDE | |
1260 : : ~ | |
1261 : : ~ | |
1262 rrrrrrrrrrrrrrrrr | |
1263 +-------+ | |
1264 The speculation is discarded ---> --->| A->1 |------>| |
1265 and an updated value is +-------+ | |
1266 retrieved : : +-------+
David Howells108b42b2006-03-31 16:00:29 +01001267
1268
Paul E. McKenney241e6662011-02-10 16:54:50 -08001269TRANSITIVITY
1270------------
1271
1272Transitivity is a deeply intuitive notion about ordering that is not
1273always provided by real computer systems. The following example
Paul E. McKenneyf36fe1e2016-02-15 14:50:36 -08001274demonstrates transitivity:
Paul E. McKenney241e6662011-02-10 16:54:50 -08001275
1276 CPU 1 CPU 2 CPU 3
1277 ======================= ======================= =======================
1278 { X = 0, Y = 0 }
1279 STORE X=1 LOAD X STORE Y=1
1280 <general barrier> <general barrier>
1281 LOAD Y LOAD X
1282
1283Suppose that CPU 2's load from X returns 1 and its load from Y returns 0.
1284This indicates that CPU 2's load from X in some sense follows CPU 1's
1285store to X and that CPU 2's load from Y in some sense preceded CPU 3's
1286store to Y. The question is then "Can CPU 3's load from X return 0?"
1287
1288Because CPU 2's load from X in some sense came after CPU 1's store, it
1289is natural to expect that CPU 3's load from X must therefore return 1.
1290This expectation is an example of transitivity: if a load executing on
1291CPU A follows a load from the same variable executing on CPU B, then
1292CPU A's load must either return the same value that CPU B's load did,
1293or must return some later value.
1294
1295In the Linux kernel, use of general memory barriers guarantees
1296transitivity. Therefore, in the above example, if CPU 2's load from X
1297returns 1 and its load from Y returns 0, then CPU 3's load from X must
1298also return 1.
1299
1300However, transitivity is -not- guaranteed for read or write barriers.
1301For example, suppose that CPU 2's general barrier in the above example
1302is changed to a read barrier as shown below:
1303
1304 CPU 1 CPU 2 CPU 3
1305 ======================= ======================= =======================
1306 { X = 0, Y = 0 }
1307 STORE X=1 LOAD X STORE Y=1
1308 <read barrier> <general barrier>
1309 LOAD Y LOAD X
1310
1311This substitution destroys transitivity: in this example, it is perfectly
1312legal for CPU 2's load from X to return 1, its load from Y to return 0,
1313and CPU 3's load from X to return 0.
1314
1315The key point is that although CPU 2's read barrier orders its pair
1316of loads, it does not guarantee to order CPU 1's store. Therefore, if
1317this example runs on a system where CPUs 1 and 2 share a store buffer
1318or a level of cache, CPU 2 might have early access to CPU 1's writes.
1319General barriers are therefore required to ensure that all CPUs agree
1320on the combined order of CPU 1's and CPU 2's accesses.
1321
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001322General barriers provide "global transitivity", so that all CPUs will
1323agree on the order of operations. In contrast, a chain of release-acquire
1324pairs provides only "local transitivity", so that only those CPUs on
1325the chain are guaranteed to agree on the combined order of the accesses.
1326For example, switching to C code in deference to Herman Hollerith:
1327
1328 int u, v, x, y, z;
1329
1330 void cpu0(void)
1331 {
1332 r0 = smp_load_acquire(&x);
1333 WRITE_ONCE(u, 1);
1334 smp_store_release(&y, 1);
1335 }
1336
1337 void cpu1(void)
1338 {
1339 r1 = smp_load_acquire(&y);
1340 r4 = READ_ONCE(v);
1341 r5 = READ_ONCE(u);
1342 smp_store_release(&z, 1);
1343 }
1344
1345 void cpu2(void)
1346 {
1347 r2 = smp_load_acquire(&z);
1348 smp_store_release(&x, 1);
1349 }
1350
1351 void cpu3(void)
1352 {
1353 WRITE_ONCE(v, 1);
1354 smp_mb();
1355 r3 = READ_ONCE(u);
1356 }
1357
1358Because cpu0(), cpu1(), and cpu2() participate in a local transitive
1359chain of smp_store_release()/smp_load_acquire() pairs, the following
1360outcome is prohibited:
1361
1362 r0 == 1 && r1 == 1 && r2 == 1
1363
1364Furthermore, because of the release-acquire relationship between cpu0()
1365and cpu1(), cpu1() must see cpu0()'s writes, so that the following
1366outcome is prohibited:
1367
1368 r1 == 1 && r5 == 0
1369
1370However, the transitivity of release-acquire is local to the participating
1371CPUs and does not apply to cpu3(). Therefore, the following outcome
1372is possible:
1373
1374 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0
1375
Paul E. McKenney37ef0342016-01-25 22:12:34 -08001376As an aside, the following outcome is also possible:
1377
1378 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0 && r5 == 1
1379
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001380Although cpu0(), cpu1(), and cpu2() will see their respective reads and
1381writes in order, CPUs not involved in the release-acquire chain might
1382well disagree on the order. This disagreement stems from the fact that
1383the weak memory-barrier instructions used to implement smp_load_acquire()
1384and smp_store_release() are not required to order prior stores against
1385subsequent loads in all cases. This means that cpu3() can see cpu0()'s
1386store to u as happening -after- cpu1()'s load from v, even though
1387both cpu0() and cpu1() agree that these two operations occurred in the
1388intended order.
1389
1390However, please keep in mind that smp_load_acquire() is not magic.
1391In particular, it simply reads from its argument with ordering. It does
1392-not- ensure that any particular value will be read. Therefore, the
1393following outcome is possible:
1394
1395 r0 == 0 && r1 == 0 && r2 == 0 && r5 == 0
1396
1397Note that this outcome can happen even on a mythical sequentially
1398consistent system where nothing is ever reordered.
1399
1400To reiterate, if your code requires global transitivity, use general
1401barriers throughout.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001402
1403
David Howells108b42b2006-03-31 16:00:29 +01001404========================
1405EXPLICIT KERNEL BARRIERS
1406========================
1407
1408The Linux kernel has a variety of different barriers that act at different
1409levels:
1410
1411 (*) Compiler barrier.
1412
1413 (*) CPU memory barriers.
1414
1415 (*) MMIO write barrier.
1416
1417
1418COMPILER BARRIER
1419----------------
1420
1421The Linux kernel has an explicit compiler barrier function that prevents the
1422compiler from moving the memory accesses either side of it to the other side:
1423
1424 barrier();
1425
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001426This is a general barrier -- there are no read-read or write-write
1427variants of barrier(). However, READ_ONCE() and WRITE_ONCE() can be
1428thought of as weak forms of barrier() that affect only the specific
1429accesses flagged by the READ_ONCE() or WRITE_ONCE().
David Howells108b42b2006-03-31 16:00:29 +01001430
Paul E. McKenney692118d2013-12-11 13:59:07 -08001431The barrier() function has the following effects:
1432
1433 (*) Prevents the compiler from reordering accesses following the
1434 barrier() to precede any accesses preceding the barrier().
1435 One example use for this property is to ease communication between
1436 interrupt-handler code and the code that was interrupted.
1437
1438 (*) Within a loop, forces the compiler to load the variables used
1439 in that loop's conditional on each pass through that loop.
1440
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001441The READ_ONCE() and WRITE_ONCE() functions can prevent any number of
1442optimizations that, while perfectly safe in single-threaded code, can
1443be fatal in concurrent code. Here are some examples of these sorts
1444of optimizations:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001445
Paul E. McKenney449f7412014-01-02 15:03:50 -08001446 (*) The compiler is within its rights to reorder loads and stores
1447 to the same variable, and in some cases, the CPU is within its
1448 rights to reorder loads to the same variable. This means that
1449 the following code:
1450
1451 a[0] = x;
1452 a[1] = x;
1453
1454 Might result in an older value of x stored in a[1] than in a[0].
1455 Prevent both the compiler and the CPU from doing this as follows:
1456
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001457 a[0] = READ_ONCE(x);
1458 a[1] = READ_ONCE(x);
Paul E. McKenney449f7412014-01-02 15:03:50 -08001459
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001460 In short, READ_ONCE() and WRITE_ONCE() provide cache coherence for
1461 accesses from multiple CPUs to a single variable.
Paul E. McKenney449f7412014-01-02 15:03:50 -08001462
Paul E. McKenney692118d2013-12-11 13:59:07 -08001463 (*) The compiler is within its rights to merge successive loads from
1464 the same variable. Such merging can cause the compiler to "optimize"
1465 the following code:
1466
1467 while (tmp = a)
1468 do_something_with(tmp);
1469
1470 into the following code, which, although in some sense legitimate
1471 for single-threaded code, is almost certainly not what the developer
1472 intended:
1473
1474 if (tmp = a)
1475 for (;;)
1476 do_something_with(tmp);
1477
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001478 Use READ_ONCE() to prevent the compiler from doing this to you:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001479
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001480 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001481 do_something_with(tmp);
1482
1483 (*) The compiler is within its rights to reload a variable, for example,
1484 in cases where high register pressure prevents the compiler from
1485 keeping all data of interest in registers. The compiler might
1486 therefore optimize the variable 'tmp' out of our previous example:
1487
1488 while (tmp = a)
1489 do_something_with(tmp);
1490
1491 This could result in the following code, which is perfectly safe in
1492 single-threaded code, but can be fatal in concurrent code:
1493
1494 while (a)
1495 do_something_with(a);
1496
1497 For example, the optimized version of this code could result in
1498 passing a zero to do_something_with() in the case where the variable
1499 a was modified by some other CPU between the "while" statement and
1500 the call to do_something_with().
1501
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001502 Again, use READ_ONCE() to prevent the compiler from doing this:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001503
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001504 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001505 do_something_with(tmp);
1506
1507 Note that if the compiler runs short of registers, it might save
1508 tmp onto the stack. The overhead of this saving and later restoring
1509 is why compilers reload variables. Doing so is perfectly safe for
1510 single-threaded code, so you need to tell the compiler about cases
1511 where it is not safe.
1512
1513 (*) The compiler is within its rights to omit a load entirely if it knows
1514 what the value will be. For example, if the compiler can prove that
1515 the value of variable 'a' is always zero, it can optimize this code:
1516
1517 while (tmp = a)
1518 do_something_with(tmp);
1519
1520 Into this:
1521
1522 do { } while (0);
1523
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001524 This transformation is a win for single-threaded code because it
1525 gets rid of a load and a branch. The problem is that the compiler
1526 will carry out its proof assuming that the current CPU is the only
1527 one updating variable 'a'. If variable 'a' is shared, then the
1528 compiler's proof will be erroneous. Use READ_ONCE() to tell the
1529 compiler that it doesn't know as much as it thinks it does:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001530
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001531 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001532 do_something_with(tmp);
1533
1534 But please note that the compiler is also closely watching what you
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001535 do with the value after the READ_ONCE(). For example, suppose you
Paul E. McKenney692118d2013-12-11 13:59:07 -08001536 do the following and MAX is a preprocessor macro with the value 1:
1537
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001538 while ((tmp = READ_ONCE(a)) % MAX)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001539 do_something_with(tmp);
1540
1541 Then the compiler knows that the result of the "%" operator applied
1542 to MAX will always be zero, again allowing the compiler to optimize
1543 the code into near-nonexistence. (It will still load from the
1544 variable 'a'.)
1545
1546 (*) Similarly, the compiler is within its rights to omit a store entirely
1547 if it knows that the variable already has the value being stored.
1548 Again, the compiler assumes that the current CPU is the only one
1549 storing into the variable, which can cause the compiler to do the
1550 wrong thing for shared variables. For example, suppose you have
1551 the following:
1552
1553 a = 0;
SeongJae Park65f95ff2016-02-22 08:28:29 -08001554 ... Code that does not store to variable a ...
Paul E. McKenney692118d2013-12-11 13:59:07 -08001555 a = 0;
1556
1557 The compiler sees that the value of variable 'a' is already zero, so
1558 it might well omit the second store. This would come as a fatal
1559 surprise if some other CPU might have stored to variable 'a' in the
1560 meantime.
1561
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001562 Use WRITE_ONCE() to prevent the compiler from making this sort of
Paul E. McKenney692118d2013-12-11 13:59:07 -08001563 wrong guess:
1564
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001565 WRITE_ONCE(a, 0);
SeongJae Park65f95ff2016-02-22 08:28:29 -08001566 ... Code that does not store to variable a ...
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001567 WRITE_ONCE(a, 0);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001568
1569 (*) The compiler is within its rights to reorder memory accesses unless
1570 you tell it not to. For example, consider the following interaction
1571 between process-level code and an interrupt handler:
1572
1573 void process_level(void)
1574 {
1575 msg = get_message();
1576 flag = true;
1577 }
1578
1579 void interrupt_handler(void)
1580 {
1581 if (flag)
1582 process_message(msg);
1583 }
1584
Masanari Iidadf5cbb22014-03-21 10:04:30 +09001585 There is nothing to prevent the compiler from transforming
Paul E. McKenney692118d2013-12-11 13:59:07 -08001586 process_level() to the following, in fact, this might well be a
1587 win for single-threaded code:
1588
1589 void process_level(void)
1590 {
1591 flag = true;
1592 msg = get_message();
1593 }
1594
1595 If the interrupt occurs between these two statement, then
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001596 interrupt_handler() might be passed a garbled msg. Use WRITE_ONCE()
Paul E. McKenney692118d2013-12-11 13:59:07 -08001597 to prevent this as follows:
1598
1599 void process_level(void)
1600 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001601 WRITE_ONCE(msg, get_message());
1602 WRITE_ONCE(flag, true);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001603 }
1604
1605 void interrupt_handler(void)
1606 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001607 if (READ_ONCE(flag))
1608 process_message(READ_ONCE(msg));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001609 }
1610
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001611 Note that the READ_ONCE() and WRITE_ONCE() wrappers in
1612 interrupt_handler() are needed if this interrupt handler can itself
1613 be interrupted by something that also accesses 'flag' and 'msg',
1614 for example, a nested interrupt or an NMI. Otherwise, READ_ONCE()
1615 and WRITE_ONCE() are not needed in interrupt_handler() other than
1616 for documentation purposes. (Note also that nested interrupts
1617 do not typically occur in modern Linux kernels, in fact, if an
1618 interrupt handler returns with interrupts enabled, you will get a
1619 WARN_ONCE() splat.)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001620
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001621 You should assume that the compiler can move READ_ONCE() and
1622 WRITE_ONCE() past code not containing READ_ONCE(), WRITE_ONCE(),
1623 barrier(), or similar primitives.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001624
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001625 This effect could also be achieved using barrier(), but READ_ONCE()
1626 and WRITE_ONCE() are more selective: With READ_ONCE() and
1627 WRITE_ONCE(), the compiler need only forget the contents of the
1628 indicated memory locations, while with barrier() the compiler must
1629 discard the value of all memory locations that it has currented
1630 cached in any machine registers. Of course, the compiler must also
1631 respect the order in which the READ_ONCE()s and WRITE_ONCE()s occur,
1632 though the CPU of course need not do so.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001633
1634 (*) The compiler is within its rights to invent stores to a variable,
1635 as in the following example:
1636
1637 if (a)
1638 b = a;
1639 else
1640 b = 42;
1641
1642 The compiler might save a branch by optimizing this as follows:
1643
1644 b = 42;
1645 if (a)
1646 b = a;
1647
1648 In single-threaded code, this is not only safe, but also saves
1649 a branch. Unfortunately, in concurrent code, this optimization
1650 could cause some other CPU to see a spurious value of 42 -- even
1651 if variable 'a' was never zero -- when loading variable 'b'.
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001652 Use WRITE_ONCE() to prevent this as follows:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001653
1654 if (a)
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001655 WRITE_ONCE(b, a);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001656 else
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001657 WRITE_ONCE(b, 42);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001658
1659 The compiler can also invent loads. These are usually less
1660 damaging, but they can result in cache-line bouncing and thus in
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001661 poor performance and scalability. Use READ_ONCE() to prevent
Paul E. McKenney692118d2013-12-11 13:59:07 -08001662 invented loads.
1663
1664 (*) For aligned memory locations whose size allows them to be accessed
1665 with a single memory-reference instruction, prevents "load tearing"
1666 and "store tearing," in which a single large access is replaced by
1667 multiple smaller accesses. For example, given an architecture having
1668 16-bit store instructions with 7-bit immediate fields, the compiler
1669 might be tempted to use two 16-bit store-immediate instructions to
1670 implement the following 32-bit store:
1671
1672 p = 0x00010002;
1673
1674 Please note that GCC really does use this sort of optimization,
1675 which is not surprising given that it would likely take more
1676 than two instructions to build the constant and then store it.
1677 This optimization can therefore be a win in single-threaded code.
1678 In fact, a recent bug (since fixed) caused GCC to incorrectly use
1679 this optimization in a volatile store. In the absence of such bugs,
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001680 use of WRITE_ONCE() prevents store tearing in the following example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001681
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001682 WRITE_ONCE(p, 0x00010002);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001683
1684 Use of packed structures can also result in load and store tearing,
1685 as in this example:
1686
1687 struct __attribute__((__packed__)) foo {
1688 short a;
1689 int b;
1690 short c;
1691 };
1692 struct foo foo1, foo2;
1693 ...
1694
1695 foo2.a = foo1.a;
1696 foo2.b = foo1.b;
1697 foo2.c = foo1.c;
1698
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001699 Because there are no READ_ONCE() or WRITE_ONCE() wrappers and no
1700 volatile markings, the compiler would be well within its rights to
1701 implement these three assignment statements as a pair of 32-bit
1702 loads followed by a pair of 32-bit stores. This would result in
1703 load tearing on 'foo1.b' and store tearing on 'foo2.b'. READ_ONCE()
1704 and WRITE_ONCE() again prevent tearing in this example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001705
1706 foo2.a = foo1.a;
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001707 WRITE_ONCE(foo2.b, READ_ONCE(foo1.b));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001708 foo2.c = foo1.c;
1709
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001710All that aside, it is never necessary to use READ_ONCE() and
1711WRITE_ONCE() on a variable that has been marked volatile. For example,
1712because 'jiffies' is marked volatile, it is never necessary to
1713say READ_ONCE(jiffies). The reason for this is that READ_ONCE() and
1714WRITE_ONCE() are implemented as volatile casts, which has no effect when
1715its argument is already marked volatile.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001716
1717Please note that these compiler barriers have no direct effect on the CPU,
1718which may then reorder things however it wishes.
David Howells108b42b2006-03-31 16:00:29 +01001719
1720
1721CPU MEMORY BARRIERS
1722-------------------
1723
1724The Linux kernel has eight basic CPU memory barriers:
1725
1726 TYPE MANDATORY SMP CONDITIONAL
1727 =============== ======================= ===========================
1728 GENERAL mb() smp_mb()
1729 WRITE wmb() smp_wmb()
1730 READ rmb() smp_rmb()
1731 DATA DEPENDENCY read_barrier_depends() smp_read_barrier_depends()
1732
1733
Nick Piggin73f10282008-05-14 06:35:11 +02001734All memory barriers except the data dependency barriers imply a compiler
1735barrier. Data dependencies do not impose any additional compiler ordering.
1736
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001737Aside: In the case of data dependencies, the compiler would be expected
1738to issue the loads in the correct order (eg. `a[b]` would have to load
1739the value of b before loading a[b]), however there is no guarantee in
1740the C specification that the compiler may not speculate the value of b
1741(eg. is equal to 1) and load a before b (eg. tmp = a[1]; if (b != 1)
1742tmp = a[b]; ). There is also the problem of a compiler reloading b after
1743having loaded a[b], thus having a newer copy of b than a[b]. A consensus
1744has not yet been reached about these problems, however the READ_ONCE()
1745macro is a good place to start looking.
David Howells108b42b2006-03-31 16:00:29 +01001746
1747SMP memory barriers are reduced to compiler barriers on uniprocessor compiled
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001748systems because it is assumed that a CPU will appear to be self-consistent,
David Howells108b42b2006-03-31 16:00:29 +01001749and will order overlapping accesses correctly with respect to itself.
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001750However, see the subsection on "Virtual Machine Guests" below.
David Howells108b42b2006-03-31 16:00:29 +01001751
1752[!] Note that SMP memory barriers _must_ be used to control the ordering of
1753references to shared memory on SMP systems, though the use of locking instead
1754is sufficient.
1755
1756Mandatory barriers should not be used to control SMP effects, since mandatory
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001757barriers impose unnecessary overhead on both SMP and UP systems. They may,
1758however, be used to control MMIO effects on accesses through relaxed memory I/O
1759windows. These barriers are required even on non-SMP systems as they affect
1760the order in which memory operations appear to a device by prohibiting both the
1761compiler and the CPU from reordering them.
David Howells108b42b2006-03-31 16:00:29 +01001762
1763
1764There are some more advanced barrier functions:
1765
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02001766 (*) smp_store_mb(var, value)
David Howells108b42b2006-03-31 16:00:29 +01001767
Oleg Nesterov75b2bd52006-11-08 17:44:38 -08001768 This assigns the value to the variable and then inserts a full memory
Davidlohr Bueso2d142e52015-10-27 12:53:51 -07001769 barrier after it. It isn't guaranteed to insert anything more than a
1770 compiler barrier in a UP compilation.
David Howells108b42b2006-03-31 16:00:29 +01001771
1772
Peter Zijlstra1b156112014-03-13 19:00:35 +01001773 (*) smp_mb__before_atomic();
1774 (*) smp_mb__after_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001775
Peter Zijlstra1b156112014-03-13 19:00:35 +01001776 These are for use with atomic (such as add, subtract, increment and
1777 decrement) functions that don't return a value, especially when used for
1778 reference counting. These functions do not imply memory barriers.
1779
1780 These are also used for atomic bitop functions that do not return a
1781 value (such as set_bit and clear_bit).
David Howells108b42b2006-03-31 16:00:29 +01001782
1783 As an example, consider a piece of code that marks an object as being dead
1784 and then decrements the object's reference count:
1785
1786 obj->dead = 1;
Peter Zijlstra1b156112014-03-13 19:00:35 +01001787 smp_mb__before_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001788 atomic_dec(&obj->ref_count);
1789
1790 This makes sure that the death mark on the object is perceived to be set
1791 *before* the reference counter is decremented.
1792
1793 See Documentation/atomic_ops.txt for more information. See the "Atomic
1794 operations" subsection for information on where to use these.
1795
1796
Paul E. McKenneyad2ad5d2015-09-17 08:18:32 -07001797 (*) lockless_dereference();
1798 This can be thought of as a pointer-fetch wrapper around the
1799 smp_read_barrier_depends() data-dependency barrier.
1800
1801 This is also similar to rcu_dereference(), but in cases where
1802 object lifetime is handled by some mechanism other than RCU, for
1803 example, when the objects removed only when the system goes down.
1804 In addition, lockless_dereference() is used in some data structures
1805 that can be used both with and without RCU.
1806
1807
Alexander Duyck1077fa32014-12-11 15:02:06 -08001808 (*) dma_wmb();
1809 (*) dma_rmb();
1810
1811 These are for use with consistent memory to guarantee the ordering
1812 of writes or reads of shared memory accessible to both the CPU and a
1813 DMA capable device.
1814
1815 For example, consider a device driver that shares memory with a device
1816 and uses a descriptor status value to indicate if the descriptor belongs
1817 to the device or the CPU, and a doorbell to notify it when new
1818 descriptors are available:
1819
1820 if (desc->status != DEVICE_OWN) {
1821 /* do not read data until we own descriptor */
1822 dma_rmb();
1823
1824 /* read/modify data */
1825 read_data = desc->data;
1826 desc->data = write_data;
1827
1828 /* flush modifications before status update */
1829 dma_wmb();
1830
1831 /* assign ownership */
1832 desc->status = DEVICE_OWN;
1833
1834 /* force memory to sync before notifying device via MMIO */
1835 wmb();
1836
1837 /* notify device of new descriptors */
1838 writel(DESC_NOTIFY, doorbell);
1839 }
1840
1841 The dma_rmb() allows us guarantee the device has released ownership
Sylvain Trias7a458002015-04-08 10:27:57 +02001842 before we read the data from the descriptor, and the dma_wmb() allows
Alexander Duyck1077fa32014-12-11 15:02:06 -08001843 us to guarantee the data is written to the descriptor before the device
1844 can see it now has ownership. The wmb() is needed to guarantee that the
1845 cache coherent memory writes have completed before attempting a write to
1846 the cache incoherent MMIO region.
1847
1848 See Documentation/DMA-API.txt for more information on consistent memory.
1849
David Howells108b42b2006-03-31 16:00:29 +01001850MMIO WRITE BARRIER
1851------------------
1852
1853The Linux kernel also has a special barrier for use with memory-mapped I/O
1854writes:
1855
1856 mmiowb();
1857
1858This is a variation on the mandatory write barrier that causes writes to weakly
1859ordered I/O regions to be partially ordered. Its effects may go beyond the
1860CPU->Hardware interface and actually affect the hardware at some level.
1861
1862See the subsection "Locks vs I/O accesses" for more information.
1863
1864
1865===============================
1866IMPLICIT KERNEL MEMORY BARRIERS
1867===============================
1868
1869Some of the other functions in the linux kernel imply memory barriers, amongst
David Howells670bd952006-06-10 09:54:12 -07001870which are locking and scheduling functions.
David Howells108b42b2006-03-31 16:00:29 +01001871
1872This specification is a _minimum_ guarantee; any particular architecture may
1873provide more substantial guarantees, but these may not be relied upon outside
1874of arch specific code.
1875
1876
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001877ACQUIRING FUNCTIONS
1878-------------------
David Howells108b42b2006-03-31 16:00:29 +01001879
1880The Linux kernel has a number of locking constructs:
1881
1882 (*) spin locks
1883 (*) R/W spin locks
1884 (*) mutexes
1885 (*) semaphores
1886 (*) R/W semaphores
David Howells108b42b2006-03-31 16:00:29 +01001887
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001888In all cases there are variants on "ACQUIRE" operations and "RELEASE" operations
David Howells108b42b2006-03-31 16:00:29 +01001889for each construct. These operations all imply certain barriers:
1890
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001891 (1) ACQUIRE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001892
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001893 Memory operations issued after the ACQUIRE will be completed after the
1894 ACQUIRE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001895
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001896 Memory operations issued before the ACQUIRE may be completed after
1897 the ACQUIRE operation has completed. An smp_mb__before_spinlock(),
Will Deacond9560282015-03-31 09:39:41 +01001898 combined with a following ACQUIRE, orders prior stores against
1899 subsequent loads and stores. Note that this is weaker than smp_mb()!
1900 The smp_mb__before_spinlock() primitive is free on many architectures.
David Howells108b42b2006-03-31 16:00:29 +01001901
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001902 (2) RELEASE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001903
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001904 Memory operations issued before the RELEASE will be completed before the
1905 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001906
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001907 Memory operations issued after the RELEASE may be completed before the
1908 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001909
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001910 (3) ACQUIRE vs ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01001911
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001912 All ACQUIRE operations issued before another ACQUIRE operation will be
1913 completed before that ACQUIRE operation.
David Howells108b42b2006-03-31 16:00:29 +01001914
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001915 (4) ACQUIRE vs RELEASE implication:
David Howells108b42b2006-03-31 16:00:29 +01001916
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001917 All ACQUIRE operations issued before a RELEASE operation will be
1918 completed before the RELEASE operation.
David Howells108b42b2006-03-31 16:00:29 +01001919
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001920 (5) Failed conditional ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01001921
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001922 Certain locking variants of the ACQUIRE operation may fail, either due to
1923 being unable to get the lock immediately, or due to receiving an unblocked
David Howells108b42b2006-03-31 16:00:29 +01001924 signal whilst asleep waiting for the lock to become available. Failed
1925 locks do not imply any sort of barrier.
1926
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001927[!] Note: one of the consequences of lock ACQUIREs and RELEASEs being only
1928one-way barriers is that the effects of instructions outside of a critical
1929section may seep into the inside of the critical section.
David Howells108b42b2006-03-31 16:00:29 +01001930
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001931An ACQUIRE followed by a RELEASE may not be assumed to be full memory barrier
1932because it is possible for an access preceding the ACQUIRE to happen after the
1933ACQUIRE, and an access following the RELEASE to happen before the RELEASE, and
1934the two accesses can themselves then cross:
David Howells670bd952006-06-10 09:54:12 -07001935
1936 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001937 ACQUIRE M
1938 RELEASE M
David Howells670bd952006-06-10 09:54:12 -07001939 *B = b;
1940
1941may occur as:
1942
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001943 ACQUIRE M, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001944
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001945When the ACQUIRE and RELEASE are a lock acquisition and release,
1946respectively, this same reordering can occur if the lock's ACQUIRE and
1947RELEASE are to the same lock variable, but only from the perspective of
1948another CPU not holding that lock. In short, a ACQUIRE followed by an
1949RELEASE may -not- be assumed to be a full memory barrier.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001950
Paul E. McKenney12d560f2015-07-14 18:35:23 -07001951Similarly, the reverse case of a RELEASE followed by an ACQUIRE does
1952not imply a full memory barrier. Therefore, the CPU's execution of the
1953critical sections corresponding to the RELEASE and the ACQUIRE can cross,
1954so that:
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001955
1956 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001957 RELEASE M
1958 ACQUIRE N
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001959 *B = b;
1960
1961could occur as:
1962
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001963 ACQUIRE N, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001964
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001965It might appear that this reordering could introduce a deadlock.
1966However, this cannot happen because if such a deadlock threatened,
1967the RELEASE would simply complete, thereby avoiding the deadlock.
1968
1969 Why does this work?
1970
1971 One key point is that we are only talking about the CPU doing
1972 the reordering, not the compiler. If the compiler (or, for
1973 that matter, the developer) switched the operations, deadlock
1974 -could- occur.
1975
1976 But suppose the CPU reordered the operations. In this case,
1977 the unlock precedes the lock in the assembly code. The CPU
1978 simply elected to try executing the later lock operation first.
1979 If there is a deadlock, this lock operation will simply spin (or
1980 try to sleep, but more on that later). The CPU will eventually
1981 execute the unlock operation (which preceded the lock operation
1982 in the assembly code), which will unravel the potential deadlock,
1983 allowing the lock operation to succeed.
1984
1985 But what if the lock is a sleeplock? In that case, the code will
1986 try to enter the scheduler, where it will eventually encounter
1987 a memory barrier, which will force the earlier unlock operation
1988 to complete, again unraveling the deadlock. There might be
1989 a sleep-unlock race, but the locking primitive needs to resolve
1990 such races properly in any case.
1991
David Howells108b42b2006-03-31 16:00:29 +01001992Locks and semaphores may not provide any guarantee of ordering on UP compiled
1993systems, and so cannot be counted on in such a situation to actually achieve
1994anything at all - especially with respect to I/O accesses - unless combined
1995with interrupt disabling operations.
1996
1997See also the section on "Inter-CPU locking barrier effects".
1998
1999
2000As an example, consider the following:
2001
2002 *A = a;
2003 *B = b;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002004 ACQUIRE
David Howells108b42b2006-03-31 16:00:29 +01002005 *C = c;
2006 *D = d;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002007 RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002008 *E = e;
2009 *F = f;
2010
2011The following sequence of events is acceptable:
2012
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002013 ACQUIRE, {*F,*A}, *E, {*C,*D}, *B, RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002014
2015 [+] Note that {*F,*A} indicates a combined access.
2016
2017But none of the following are:
2018
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002019 {*F,*A}, *B, ACQUIRE, *C, *D, RELEASE, *E
2020 *A, *B, *C, ACQUIRE, *D, RELEASE, *E, *F
2021 *A, *B, ACQUIRE, *C, RELEASE, *D, *E, *F
2022 *B, ACQUIRE, *C, *D, RELEASE, {*F,*A}, *E
David Howells108b42b2006-03-31 16:00:29 +01002023
2024
2025
2026INTERRUPT DISABLING FUNCTIONS
2027-----------------------------
2028
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002029Functions that disable interrupts (ACQUIRE equivalent) and enable interrupts
2030(RELEASE equivalent) will act as compiler barriers only. So if memory or I/O
David Howells108b42b2006-03-31 16:00:29 +01002031barriers are required in such a situation, they must be provided from some
2032other means.
2033
2034
David Howells50fa6102009-04-28 15:01:38 +01002035SLEEP AND WAKE-UP FUNCTIONS
2036---------------------------
2037
2038Sleeping and waking on an event flagged in global data can be viewed as an
2039interaction between two pieces of data: the task state of the task waiting for
2040the event and the global data used to indicate the event. To make sure that
2041these appear to happen in the right order, the primitives to begin the process
2042of going to sleep, and the primitives to initiate a wake up imply certain
2043barriers.
2044
2045Firstly, the sleeper normally follows something like this sequence of events:
2046
2047 for (;;) {
2048 set_current_state(TASK_UNINTERRUPTIBLE);
2049 if (event_indicated)
2050 break;
2051 schedule();
2052 }
2053
2054A general memory barrier is interpolated automatically by set_current_state()
2055after it has altered the task state:
2056
2057 CPU 1
2058 ===============================
2059 set_current_state();
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002060 smp_store_mb();
David Howells50fa6102009-04-28 15:01:38 +01002061 STORE current->state
2062 <general barrier>
2063 LOAD event_indicated
2064
2065set_current_state() may be wrapped by:
2066
2067 prepare_to_wait();
2068 prepare_to_wait_exclusive();
2069
2070which therefore also imply a general memory barrier after setting the state.
2071The whole sequence above is available in various canned forms, all of which
2072interpolate the memory barrier in the right place:
2073
2074 wait_event();
2075 wait_event_interruptible();
2076 wait_event_interruptible_exclusive();
2077 wait_event_interruptible_timeout();
2078 wait_event_killable();
2079 wait_event_timeout();
2080 wait_on_bit();
2081 wait_on_bit_lock();
2082
2083
2084Secondly, code that performs a wake up normally follows something like this:
2085
2086 event_indicated = 1;
2087 wake_up(&event_wait_queue);
2088
2089or:
2090
2091 event_indicated = 1;
2092 wake_up_process(event_daemon);
2093
2094A write memory barrier is implied by wake_up() and co. if and only if they wake
2095something up. The barrier occurs before the task state is cleared, and so sits
2096between the STORE to indicate the event and the STORE to set TASK_RUNNING:
2097
2098 CPU 1 CPU 2
2099 =============================== ===============================
2100 set_current_state(); STORE event_indicated
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002101 smp_store_mb(); wake_up();
David Howells50fa6102009-04-28 15:01:38 +01002102 STORE current->state <write barrier>
2103 <general barrier> STORE current->state
2104 LOAD event_indicated
2105
Paul E. McKenney5726ce02014-05-13 10:14:51 -07002106To repeat, this write memory barrier is present if and only if something
2107is actually awakened. To see this, consider the following sequence of
2108events, where X and Y are both initially zero:
2109
2110 CPU 1 CPU 2
2111 =============================== ===============================
2112 X = 1; STORE event_indicated
2113 smp_mb(); wake_up();
2114 Y = 1; wait_event(wq, Y == 1);
2115 wake_up(); load from Y sees 1, no memory barrier
2116 load from X might see 0
2117
2118In contrast, if a wakeup does occur, CPU 2's load from X would be guaranteed
2119to see 1.
2120
David Howells50fa6102009-04-28 15:01:38 +01002121The available waker functions include:
2122
2123 complete();
2124 wake_up();
2125 wake_up_all();
2126 wake_up_bit();
2127 wake_up_interruptible();
2128 wake_up_interruptible_all();
2129 wake_up_interruptible_nr();
2130 wake_up_interruptible_poll();
2131 wake_up_interruptible_sync();
2132 wake_up_interruptible_sync_poll();
2133 wake_up_locked();
2134 wake_up_locked_poll();
2135 wake_up_nr();
2136 wake_up_poll();
2137 wake_up_process();
2138
2139
2140[!] Note that the memory barriers implied by the sleeper and the waker do _not_
2141order multiple stores before the wake-up with respect to loads of those stored
2142values after the sleeper has called set_current_state(). For instance, if the
2143sleeper does:
2144
2145 set_current_state(TASK_INTERRUPTIBLE);
2146 if (event_indicated)
2147 break;
2148 __set_current_state(TASK_RUNNING);
2149 do_something(my_data);
2150
2151and the waker does:
2152
2153 my_data = value;
2154 event_indicated = 1;
2155 wake_up(&event_wait_queue);
2156
2157there's no guarantee that the change to event_indicated will be perceived by
2158the sleeper as coming after the change to my_data. In such a circumstance, the
2159code on both sides must interpolate its own memory barriers between the
2160separate data accesses. Thus the above sleeper ought to do:
2161
2162 set_current_state(TASK_INTERRUPTIBLE);
2163 if (event_indicated) {
2164 smp_rmb();
2165 do_something(my_data);
2166 }
2167
2168and the waker should do:
2169
2170 my_data = value;
2171 smp_wmb();
2172 event_indicated = 1;
2173 wake_up(&event_wait_queue);
2174
2175
David Howells108b42b2006-03-31 16:00:29 +01002176MISCELLANEOUS FUNCTIONS
2177-----------------------
2178
2179Other functions that imply barriers:
2180
2181 (*) schedule() and similar imply full memory barriers.
2182
David Howells108b42b2006-03-31 16:00:29 +01002183
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002184===================================
2185INTER-CPU ACQUIRING BARRIER EFFECTS
2186===================================
David Howells108b42b2006-03-31 16:00:29 +01002187
2188On SMP systems locking primitives give a more substantial form of barrier: one
2189that does affect memory access ordering on other CPUs, within the context of
2190conflict on any particular lock.
2191
2192
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002193ACQUIRES VS MEMORY ACCESSES
2194---------------------------
David Howells108b42b2006-03-31 16:00:29 +01002195
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002196Consider the following: the system has a pair of spinlocks (M) and (Q), and
David Howells108b42b2006-03-31 16:00:29 +01002197three CPUs; then should the following sequence of events occur:
2198
2199 CPU 1 CPU 2
2200 =============================== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002201 WRITE_ONCE(*A, a); WRITE_ONCE(*E, e);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002202 ACQUIRE M ACQUIRE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002203 WRITE_ONCE(*B, b); WRITE_ONCE(*F, f);
2204 WRITE_ONCE(*C, c); WRITE_ONCE(*G, g);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002205 RELEASE M RELEASE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002206 WRITE_ONCE(*D, d); WRITE_ONCE(*H, h);
David Howells108b42b2006-03-31 16:00:29 +01002207
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002208Then there is no guarantee as to what order CPU 3 will see the accesses to *A
David Howells108b42b2006-03-31 16:00:29 +01002209through *H occur in, other than the constraints imposed by the separate locks
2210on the separate CPUs. It might, for example, see:
2211
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002212 *E, ACQUIRE M, ACQUIRE Q, *G, *C, *F, *A, *B, RELEASE Q, *D, *H, RELEASE M
David Howells108b42b2006-03-31 16:00:29 +01002213
2214But it won't see any of:
2215
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002216 *B, *C or *D preceding ACQUIRE M
2217 *A, *B or *C following RELEASE M
2218 *F, *G or *H preceding ACQUIRE Q
2219 *E, *F or *G following RELEASE Q
David Howells108b42b2006-03-31 16:00:29 +01002220
2221
David Howells108b42b2006-03-31 16:00:29 +01002222
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002223ACQUIRES VS I/O ACCESSES
2224------------------------
David Howells108b42b2006-03-31 16:00:29 +01002225
2226Under certain circumstances (especially involving NUMA), I/O accesses within
2227two spinlocked sections on two different CPUs may be seen as interleaved by the
2228PCI bridge, because the PCI bridge does not necessarily participate in the
2229cache-coherence protocol, and is therefore incapable of issuing the required
2230read memory barriers.
2231
2232For example:
2233
2234 CPU 1 CPU 2
2235 =============================== ===============================
2236 spin_lock(Q)
2237 writel(0, ADDR)
2238 writel(1, DATA);
2239 spin_unlock(Q);
2240 spin_lock(Q);
2241 writel(4, ADDR);
2242 writel(5, DATA);
2243 spin_unlock(Q);
2244
2245may be seen by the PCI bridge as follows:
2246
2247 STORE *ADDR = 0, STORE *ADDR = 4, STORE *DATA = 1, STORE *DATA = 5
2248
2249which would probably cause the hardware to malfunction.
2250
2251
2252What is necessary here is to intervene with an mmiowb() before dropping the
2253spinlock, for example:
2254
2255 CPU 1 CPU 2
2256 =============================== ===============================
2257 spin_lock(Q)
2258 writel(0, ADDR)
2259 writel(1, DATA);
2260 mmiowb();
2261 spin_unlock(Q);
2262 spin_lock(Q);
2263 writel(4, ADDR);
2264 writel(5, DATA);
2265 mmiowb();
2266 spin_unlock(Q);
2267
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002268this will ensure that the two stores issued on CPU 1 appear at the PCI bridge
2269before either of the stores issued on CPU 2.
David Howells108b42b2006-03-31 16:00:29 +01002270
2271
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002272Furthermore, following a store by a load from the same device obviates the need
2273for the mmiowb(), because the load forces the store to complete before the load
David Howells108b42b2006-03-31 16:00:29 +01002274is performed:
2275
2276 CPU 1 CPU 2
2277 =============================== ===============================
2278 spin_lock(Q)
2279 writel(0, ADDR)
2280 a = readl(DATA);
2281 spin_unlock(Q);
2282 spin_lock(Q);
2283 writel(4, ADDR);
2284 b = readl(DATA);
2285 spin_unlock(Q);
2286
2287
2288See Documentation/DocBook/deviceiobook.tmpl for more information.
2289
2290
2291=================================
2292WHERE ARE MEMORY BARRIERS NEEDED?
2293=================================
2294
2295Under normal operation, memory operation reordering is generally not going to
2296be a problem as a single-threaded linear piece of code will still appear to
David Howells50fa6102009-04-28 15:01:38 +01002297work correctly, even if it's in an SMP kernel. There are, however, four
David Howells108b42b2006-03-31 16:00:29 +01002298circumstances in which reordering definitely _could_ be a problem:
2299
2300 (*) Interprocessor interaction.
2301
2302 (*) Atomic operations.
2303
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002304 (*) Accessing devices.
David Howells108b42b2006-03-31 16:00:29 +01002305
2306 (*) Interrupts.
2307
2308
2309INTERPROCESSOR INTERACTION
2310--------------------------
2311
2312When there's a system with more than one processor, more than one CPU in the
2313system may be working on the same data set at the same time. This can cause
2314synchronisation problems, and the usual way of dealing with them is to use
2315locks. Locks, however, are quite expensive, and so it may be preferable to
2316operate without the use of a lock if at all possible. In such a case
2317operations that affect both CPUs may have to be carefully ordered to prevent
2318a malfunction.
2319
2320Consider, for example, the R/W semaphore slow path. Here a waiting process is
2321queued on the semaphore, by virtue of it having a piece of its stack linked to
2322the semaphore's list of waiting processes:
2323
2324 struct rw_semaphore {
2325 ...
2326 spinlock_t lock;
2327 struct list_head waiters;
2328 };
2329
2330 struct rwsem_waiter {
2331 struct list_head list;
2332 struct task_struct *task;
2333 };
2334
2335To wake up a particular waiter, the up_read() or up_write() functions have to:
2336
2337 (1) read the next pointer from this waiter's record to know as to where the
2338 next waiter record is;
2339
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002340 (2) read the pointer to the waiter's task structure;
David Howells108b42b2006-03-31 16:00:29 +01002341
2342 (3) clear the task pointer to tell the waiter it has been given the semaphore;
2343
2344 (4) call wake_up_process() on the task; and
2345
2346 (5) release the reference held on the waiter's task struct.
2347
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002348In other words, it has to perform this sequence of events:
David Howells108b42b2006-03-31 16:00:29 +01002349
2350 LOAD waiter->list.next;
2351 LOAD waiter->task;
2352 STORE waiter->task;
2353 CALL wakeup
2354 RELEASE task
2355
2356and if any of these steps occur out of order, then the whole thing may
2357malfunction.
2358
2359Once it has queued itself and dropped the semaphore lock, the waiter does not
2360get the lock again; it instead just waits for its task pointer to be cleared
2361before proceeding. Since the record is on the waiter's stack, this means that
2362if the task pointer is cleared _before_ the next pointer in the list is read,
2363another CPU might start processing the waiter and might clobber the waiter's
2364stack before the up*() function has a chance to read the next pointer.
2365
2366Consider then what might happen to the above sequence of events:
2367
2368 CPU 1 CPU 2
2369 =============================== ===============================
2370 down_xxx()
2371 Queue waiter
2372 Sleep
2373 up_yyy()
2374 LOAD waiter->task;
2375 STORE waiter->task;
2376 Woken up by other event
2377 <preempt>
2378 Resume processing
2379 down_xxx() returns
2380 call foo()
2381 foo() clobbers *waiter
2382 </preempt>
2383 LOAD waiter->list.next;
2384 --- OOPS ---
2385
2386This could be dealt with using the semaphore lock, but then the down_xxx()
2387function has to needlessly get the spinlock again after being woken up.
2388
2389The way to deal with this is to insert a general SMP memory barrier:
2390
2391 LOAD waiter->list.next;
2392 LOAD waiter->task;
2393 smp_mb();
2394 STORE waiter->task;
2395 CALL wakeup
2396 RELEASE task
2397
2398In this case, the barrier makes a guarantee that all memory accesses before the
2399barrier will appear to happen before all the memory accesses after the barrier
2400with respect to the other CPUs on the system. It does _not_ guarantee that all
2401the memory accesses before the barrier will be complete by the time the barrier
2402instruction itself is complete.
2403
2404On a UP system - where this wouldn't be a problem - the smp_mb() is just a
2405compiler barrier, thus making sure the compiler emits the instructions in the
David Howells6bc39272006-06-25 05:49:22 -07002406right order without actually intervening in the CPU. Since there's only one
2407CPU, that CPU's dependency ordering logic will take care of everything else.
David Howells108b42b2006-03-31 16:00:29 +01002408
2409
2410ATOMIC OPERATIONS
2411-----------------
2412
David Howellsdbc87002006-04-10 22:54:23 -07002413Whilst they are technically interprocessor interaction considerations, atomic
2414operations are noted specially as some of them imply full memory barriers and
2415some don't, but they're very heavily relied on as a group throughout the
2416kernel.
2417
2418Any atomic operation that modifies some state in memory and returns information
2419about the state (old or new) implies an SMP-conditional general memory barrier
Nick Piggin26333572007-10-18 03:06:39 -07002420(smp_mb()) on each side of the actual operation (with the exception of
2421explicit lock operations, described later). These include:
David Howells108b42b2006-03-31 16:00:29 +01002422
2423 xchg();
Paul E. McKenneyfb2b5812013-12-11 13:59:05 -08002424 atomic_xchg(); atomic_long_xchg();
Paul E. McKenneyfb2b5812013-12-11 13:59:05 -08002425 atomic_inc_return(); atomic_long_inc_return();
2426 atomic_dec_return(); atomic_long_dec_return();
2427 atomic_add_return(); atomic_long_add_return();
2428 atomic_sub_return(); atomic_long_sub_return();
2429 atomic_inc_and_test(); atomic_long_inc_and_test();
2430 atomic_dec_and_test(); atomic_long_dec_and_test();
2431 atomic_sub_and_test(); atomic_long_sub_and_test();
2432 atomic_add_negative(); atomic_long_add_negative();
David Howellsdbc87002006-04-10 22:54:23 -07002433 test_and_set_bit();
2434 test_and_clear_bit();
2435 test_and_change_bit();
David Howells108b42b2006-03-31 16:00:29 +01002436
Will Deaconed2de9f2015-07-16 16:10:06 +01002437 /* when succeeds */
2438 cmpxchg();
2439 atomic_cmpxchg(); atomic_long_cmpxchg();
Paul E. McKenneyfb2b5812013-12-11 13:59:05 -08002440 atomic_add_unless(); atomic_long_add_unless();
2441
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002442These are used for such things as implementing ACQUIRE-class and RELEASE-class
David Howellsdbc87002006-04-10 22:54:23 -07002443operations and adjusting reference counters towards object destruction, and as
2444such the implicit memory barrier effects are necessary.
David Howells108b42b2006-03-31 16:00:29 +01002445
David Howells108b42b2006-03-31 16:00:29 +01002446
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002447The following operations are potential problems as they do _not_ imply memory
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002448barriers, but might be used for implementing such things as RELEASE-class
David Howellsdbc87002006-04-10 22:54:23 -07002449operations:
2450
2451 atomic_set();
David Howells108b42b2006-03-31 16:00:29 +01002452 set_bit();
2453 clear_bit();
2454 change_bit();
David Howellsdbc87002006-04-10 22:54:23 -07002455
2456With these the appropriate explicit memory barrier should be used if necessary
Peter Zijlstra1b156112014-03-13 19:00:35 +01002457(smp_mb__before_atomic() for instance).
David Howells108b42b2006-03-31 16:00:29 +01002458
2459
David Howellsdbc87002006-04-10 22:54:23 -07002460The following also do _not_ imply memory barriers, and so may require explicit
Peter Zijlstra1b156112014-03-13 19:00:35 +01002461memory barriers under some circumstances (smp_mb__before_atomic() for
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002462instance):
David Howells108b42b2006-03-31 16:00:29 +01002463
2464 atomic_add();
2465 atomic_sub();
2466 atomic_inc();
2467 atomic_dec();
2468
2469If they're used for statistics generation, then they probably don't need memory
2470barriers, unless there's a coupling between statistical data.
2471
2472If they're used for reference counting on an object to control its lifetime,
2473they probably don't need memory barriers because either the reference count
2474will be adjusted inside a locked section, or the caller will already hold
2475sufficient references to make the lock, and thus a memory barrier unnecessary.
2476
2477If they're used for constructing a lock of some description, then they probably
2478do need memory barriers as a lock primitive generally has to do things in a
2479specific order.
2480
David Howells108b42b2006-03-31 16:00:29 +01002481Basically, each usage case has to be carefully considered as to whether memory
David Howellsdbc87002006-04-10 22:54:23 -07002482barriers are needed or not.
2483
Nick Piggin26333572007-10-18 03:06:39 -07002484The following operations are special locking primitives:
2485
2486 test_and_set_bit_lock();
2487 clear_bit_unlock();
2488 __clear_bit_unlock();
2489
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002490These implement ACQUIRE-class and RELEASE-class operations. These should be used in
Nick Piggin26333572007-10-18 03:06:39 -07002491preference to other operations when implementing locking primitives, because
2492their implementations can be optimised on many architectures.
2493
David Howellsdbc87002006-04-10 22:54:23 -07002494[!] Note that special memory barrier primitives are available for these
2495situations because on some CPUs the atomic instructions used imply full memory
2496barriers, and so barrier instructions are superfluous in conjunction with them,
2497and in such cases the special barrier primitives will be no-ops.
David Howells108b42b2006-03-31 16:00:29 +01002498
2499See Documentation/atomic_ops.txt for more information.
2500
2501
2502ACCESSING DEVICES
2503-----------------
2504
2505Many devices can be memory mapped, and so appear to the CPU as if they're just
2506a set of memory locations. To control such a device, the driver usually has to
2507make the right memory accesses in exactly the right order.
2508
2509However, having a clever CPU or a clever compiler creates a potential problem
2510in that the carefully sequenced accesses in the driver code won't reach the
2511device in the requisite order if the CPU or the compiler thinks it is more
2512efficient to reorder, combine or merge accesses - something that would cause
2513the device to malfunction.
2514
2515Inside of the Linux kernel, I/O should be done through the appropriate accessor
2516routines - such as inb() or writel() - which know how to make such accesses
2517appropriately sequential. Whilst this, for the most part, renders the explicit
2518use of memory barriers unnecessary, there are a couple of situations where they
2519might be needed:
2520
2521 (1) On some systems, I/O stores are not strongly ordered across all CPUs, and
2522 so for _all_ general drivers locks should be used and mmiowb() must be
2523 issued prior to unlocking the critical section.
2524
2525 (2) If the accessor functions are used to refer to an I/O memory window with
2526 relaxed memory access properties, then _mandatory_ memory barriers are
2527 required to enforce ordering.
2528
2529See Documentation/DocBook/deviceiobook.tmpl for more information.
2530
2531
2532INTERRUPTS
2533----------
2534
2535A driver may be interrupted by its own interrupt service routine, and thus the
2536two parts of the driver may interfere with each other's attempts to control or
2537access the device.
2538
2539This may be alleviated - at least in part - by disabling local interrupts (a
2540form of locking), such that the critical operations are all contained within
2541the interrupt-disabled section in the driver. Whilst the driver's interrupt
2542routine is executing, the driver's core may not run on the same CPU, and its
2543interrupt is not permitted to happen again until the current interrupt has been
2544handled, thus the interrupt handler does not need to lock against that.
2545
2546However, consider a driver that was talking to an ethernet card that sports an
2547address register and a data register. If that driver's core talks to the card
2548under interrupt-disablement and then the driver's interrupt handler is invoked:
2549
2550 LOCAL IRQ DISABLE
2551 writew(ADDR, 3);
2552 writew(DATA, y);
2553 LOCAL IRQ ENABLE
2554 <interrupt>
2555 writew(ADDR, 4);
2556 q = readw(DATA);
2557 </interrupt>
2558
2559The store to the data register might happen after the second store to the
2560address register if ordering rules are sufficiently relaxed:
2561
2562 STORE *ADDR = 3, STORE *ADDR = 4, STORE *DATA = y, q = LOAD *DATA
2563
2564
2565If ordering rules are relaxed, it must be assumed that accesses done inside an
2566interrupt disabled section may leak outside of it and may interleave with
2567accesses performed in an interrupt - and vice versa - unless implicit or
2568explicit barriers are used.
2569
2570Normally this won't be a problem because the I/O accesses done inside such
2571sections will include synchronous load operations on strictly ordered I/O
2572registers that form implicit I/O barriers. If this isn't sufficient then an
2573mmiowb() may need to be used explicitly.
2574
2575
2576A similar situation may occur between an interrupt routine and two routines
2577running on separate CPUs that communicate with each other. If such a case is
2578likely, then interrupt-disabling locks should be used to guarantee ordering.
2579
2580
2581==========================
2582KERNEL I/O BARRIER EFFECTS
2583==========================
2584
2585When accessing I/O memory, drivers should use the appropriate accessor
2586functions:
2587
2588 (*) inX(), outX():
2589
2590 These are intended to talk to I/O space rather than memory space, but
2591 that's primarily a CPU-specific concept. The i386 and x86_64 processors do
2592 indeed have special I/O space access cycles and instructions, but many
2593 CPUs don't have such a concept.
2594
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002595 The PCI bus, amongst others, defines an I/O space concept which - on such
2596 CPUs as i386 and x86_64 - readily maps to the CPU's concept of I/O
David Howells6bc39272006-06-25 05:49:22 -07002597 space. However, it may also be mapped as a virtual I/O space in the CPU's
2598 memory map, particularly on those CPUs that don't support alternate I/O
2599 spaces.
David Howells108b42b2006-03-31 16:00:29 +01002600
2601 Accesses to this space may be fully synchronous (as on i386), but
2602 intermediary bridges (such as the PCI host bridge) may not fully honour
2603 that.
2604
2605 They are guaranteed to be fully ordered with respect to each other.
2606
2607 They are not guaranteed to be fully ordered with respect to other types of
2608 memory and I/O operation.
2609
2610 (*) readX(), writeX():
2611
2612 Whether these are guaranteed to be fully ordered and uncombined with
2613 respect to each other on the issuing CPU depends on the characteristics
2614 defined for the memory window through which they're accessing. On later
2615 i386 architecture machines, for example, this is controlled by way of the
2616 MTRR registers.
2617
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002618 Ordinarily, these will be guaranteed to be fully ordered and uncombined,
David Howells108b42b2006-03-31 16:00:29 +01002619 provided they're not accessing a prefetchable device.
2620
2621 However, intermediary hardware (such as a PCI bridge) may indulge in
2622 deferral if it so wishes; to flush a store, a load from the same location
2623 is preferred[*], but a load from the same device or from configuration
2624 space should suffice for PCI.
2625
2626 [*] NOTE! attempting to load from the same location as was written to may
Ingo Molnare0edc782013-11-22 11:24:53 +01002627 cause a malfunction - consider the 16550 Rx/Tx serial registers for
2628 example.
David Howells108b42b2006-03-31 16:00:29 +01002629
2630 Used with prefetchable I/O memory, an mmiowb() barrier may be required to
2631 force stores to be ordered.
2632
2633 Please refer to the PCI specification for more information on interactions
2634 between PCI transactions.
2635
Will Deacona8e0aea2013-09-04 12:30:08 +01002636 (*) readX_relaxed(), writeX_relaxed()
David Howells108b42b2006-03-31 16:00:29 +01002637
Will Deacona8e0aea2013-09-04 12:30:08 +01002638 These are similar to readX() and writeX(), but provide weaker memory
2639 ordering guarantees. Specifically, they do not guarantee ordering with
2640 respect to normal memory accesses (e.g. DMA buffers) nor do they guarantee
2641 ordering with respect to LOCK or UNLOCK operations. If the latter is
2642 required, an mmiowb() barrier can be used. Note that relaxed accesses to
2643 the same peripheral are guaranteed to be ordered with respect to each
2644 other.
David Howells108b42b2006-03-31 16:00:29 +01002645
2646 (*) ioreadX(), iowriteX()
2647
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002648 These will perform appropriately for the type of access they're actually
David Howells108b42b2006-03-31 16:00:29 +01002649 doing, be it inX()/outX() or readX()/writeX().
2650
2651
2652========================================
2653ASSUMED MINIMUM EXECUTION ORDERING MODEL
2654========================================
2655
2656It has to be assumed that the conceptual CPU is weakly-ordered but that it will
2657maintain the appearance of program causality with respect to itself. Some CPUs
2658(such as i386 or x86_64) are more constrained than others (such as powerpc or
2659frv), and so the most relaxed case (namely DEC Alpha) must be assumed outside
2660of arch-specific code.
2661
2662This means that it must be considered that the CPU will execute its instruction
2663stream in any order it feels like - or even in parallel - provided that if an
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002664instruction in the stream depends on an earlier instruction, then that
David Howells108b42b2006-03-31 16:00:29 +01002665earlier instruction must be sufficiently complete[*] before the later
2666instruction may proceed; in other words: provided that the appearance of
2667causality is maintained.
2668
2669 [*] Some instructions have more than one effect - such as changing the
2670 condition codes, changing registers or changing memory - and different
2671 instructions may depend on different effects.
2672
2673A CPU may also discard any instruction sequence that winds up having no
2674ultimate effect. For example, if two adjacent instructions both load an
2675immediate value into the same register, the first may be discarded.
2676
2677
2678Similarly, it has to be assumed that compiler might reorder the instruction
2679stream in any way it sees fit, again provided the appearance of causality is
2680maintained.
2681
2682
2683============================
2684THE EFFECTS OF THE CPU CACHE
2685============================
2686
2687The way cached memory operations are perceived across the system is affected to
2688a certain extent by the caches that lie between CPUs and memory, and by the
2689memory coherence system that maintains the consistency of state in the system.
2690
2691As far as the way a CPU interacts with another part of the system through the
2692caches goes, the memory system has to include the CPU's caches, and memory
2693barriers for the most part act at the interface between the CPU and its cache
2694(memory barriers logically act on the dotted line in the following diagram):
2695
2696 <--- CPU ---> : <----------- Memory ----------->
2697 :
2698 +--------+ +--------+ : +--------+ +-----------+
2699 | | | | : | | | | +--------+
Ingo Molnare0edc782013-11-22 11:24:53 +01002700 | CPU | | Memory | : | CPU | | | | |
2701 | Core |--->| Access |----->| Cache |<-->| | | |
David Howells108b42b2006-03-31 16:00:29 +01002702 | | | Queue | : | | | |--->| Memory |
Ingo Molnare0edc782013-11-22 11:24:53 +01002703 | | | | : | | | | | |
2704 +--------+ +--------+ : +--------+ | | | |
David Howells108b42b2006-03-31 16:00:29 +01002705 : | Cache | +--------+
2706 : | Coherency |
2707 : | Mechanism | +--------+
2708 +--------+ +--------+ : +--------+ | | | |
2709 | | | | : | | | | | |
2710 | CPU | | Memory | : | CPU | | |--->| Device |
Ingo Molnare0edc782013-11-22 11:24:53 +01002711 | Core |--->| Access |----->| Cache |<-->| | | |
2712 | | | Queue | : | | | | | |
David Howells108b42b2006-03-31 16:00:29 +01002713 | | | | : | | | | +--------+
2714 +--------+ +--------+ : +--------+ +-----------+
2715 :
2716 :
2717
2718Although any particular load or store may not actually appear outside of the
2719CPU that issued it since it may have been satisfied within the CPU's own cache,
2720it will still appear as if the full memory access had taken place as far as the
2721other CPUs are concerned since the cache coherency mechanisms will migrate the
2722cacheline over to the accessing CPU and propagate the effects upon conflict.
2723
2724The CPU core may execute instructions in any order it deems fit, provided the
2725expected program causality appears to be maintained. Some of the instructions
2726generate load and store operations which then go into the queue of memory
2727accesses to be performed. The core may place these in the queue in any order
2728it wishes, and continue execution until it is forced to wait for an instruction
2729to complete.
2730
2731What memory barriers are concerned with is controlling the order in which
2732accesses cross from the CPU side of things to the memory side of things, and
2733the order in which the effects are perceived to happen by the other observers
2734in the system.
2735
2736[!] Memory barriers are _not_ needed within a given CPU, as CPUs always see
2737their own loads and stores as if they had happened in program order.
2738
2739[!] MMIO or other device accesses may bypass the cache system. This depends on
2740the properties of the memory window through which devices are accessed and/or
2741the use of any special device communication instructions the CPU may have.
2742
2743
2744CACHE COHERENCY
2745---------------
2746
2747Life isn't quite as simple as it may appear above, however: for while the
2748caches are expected to be coherent, there's no guarantee that that coherency
2749will be ordered. This means that whilst changes made on one CPU will
2750eventually become visible on all CPUs, there's no guarantee that they will
2751become apparent in the same order on those other CPUs.
2752
2753
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002754Consider dealing with a system that has a pair of CPUs (1 & 2), each of which
2755has a pair of parallel data caches (CPU 1 has A/B, and CPU 2 has C/D):
David Howells108b42b2006-03-31 16:00:29 +01002756
2757 :
2758 : +--------+
2759 : +---------+ | |
2760 +--------+ : +--->| Cache A |<------->| |
2761 | | : | +---------+ | |
2762 | CPU 1 |<---+ | |
2763 | | : | +---------+ | |
2764 +--------+ : +--->| Cache B |<------->| |
2765 : +---------+ | |
2766 : | Memory |
2767 : +---------+ | System |
2768 +--------+ : +--->| Cache C |<------->| |
2769 | | : | +---------+ | |
2770 | CPU 2 |<---+ | |
2771 | | : | +---------+ | |
2772 +--------+ : +--->| Cache D |<------->| |
2773 : +---------+ | |
2774 : +--------+
2775 :
2776
2777Imagine the system has the following properties:
2778
2779 (*) an odd-numbered cache line may be in cache A, cache C or it may still be
2780 resident in memory;
2781
2782 (*) an even-numbered cache line may be in cache B, cache D or it may still be
2783 resident in memory;
2784
2785 (*) whilst the CPU core is interrogating one cache, the other cache may be
2786 making use of the bus to access the rest of the system - perhaps to
2787 displace a dirty cacheline or to do a speculative load;
2788
2789 (*) each cache has a queue of operations that need to be applied to that cache
2790 to maintain coherency with the rest of the system;
2791
2792 (*) the coherency queue is not flushed by normal loads to lines already
2793 present in the cache, even though the contents of the queue may
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002794 potentially affect those loads.
David Howells108b42b2006-03-31 16:00:29 +01002795
2796Imagine, then, that two writes are made on the first CPU, with a write barrier
2797between them to guarantee that they will appear to reach that CPU's caches in
2798the requisite order:
2799
2800 CPU 1 CPU 2 COMMENT
2801 =============== =============== =======================================
2802 u == 0, v == 1 and p == &u, q == &u
2803 v = 2;
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002804 smp_wmb(); Make sure change to v is visible before
David Howells108b42b2006-03-31 16:00:29 +01002805 change to p
2806 <A:modify v=2> v is now in cache A exclusively
2807 p = &v;
2808 <B:modify p=&v> p is now in cache B exclusively
2809
2810The write memory barrier forces the other CPUs in the system to perceive that
2811the local CPU's caches have apparently been updated in the correct order. But
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002812now imagine that the second CPU wants to read those values:
David Howells108b42b2006-03-31 16:00:29 +01002813
2814 CPU 1 CPU 2 COMMENT
2815 =============== =============== =======================================
2816 ...
2817 q = p;
2818 x = *q;
2819
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002820The above pair of reads may then fail to happen in the expected order, as the
David Howells108b42b2006-03-31 16:00:29 +01002821cacheline holding p may get updated in one of the second CPU's caches whilst
2822the update to the cacheline holding v is delayed in the other of the second
2823CPU's caches by some other cache event:
2824
2825 CPU 1 CPU 2 COMMENT
2826 =============== =============== =======================================
2827 u == 0, v == 1 and p == &u, q == &u
2828 v = 2;
2829 smp_wmb();
2830 <A:modify v=2> <C:busy>
2831 <C:queue v=2>
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002832 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002833 <D:request p>
2834 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002835 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002836 x = *q;
2837 <C:read *q> Reads from v before v updated in cache
2838 <C:unbusy>
2839 <C:commit v=2>
2840
2841Basically, whilst both cachelines will be updated on CPU 2 eventually, there's
2842no guarantee that, without intervention, the order of update will be the same
2843as that committed on CPU 1.
2844
2845
2846To intervene, we need to interpolate a data dependency barrier or a read
2847barrier between the loads. This will force the cache to commit its coherency
2848queue before processing any further requests:
2849
2850 CPU 1 CPU 2 COMMENT
2851 =============== =============== =======================================
2852 u == 0, v == 1 and p == &u, q == &u
2853 v = 2;
2854 smp_wmb();
2855 <A:modify v=2> <C:busy>
2856 <C:queue v=2>
Paolo 'Blaisorblade' Giarrusso3fda9822006-10-19 23:28:19 -07002857 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002858 <D:request p>
2859 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002860 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002861 smp_read_barrier_depends()
2862 <C:unbusy>
2863 <C:commit v=2>
2864 x = *q;
2865 <C:read *q> Reads from v after v updated in cache
2866
2867
2868This sort of problem can be encountered on DEC Alpha processors as they have a
2869split cache that improves performance by making better use of the data bus.
2870Whilst most CPUs do imply a data dependency barrier on the read when a memory
2871access depends on a read, not all do, so it may not be relied on.
2872
2873Other CPUs may also have split caches, but must coordinate between the various
Matt LaPlante3f6dee92006-10-03 22:45:33 +02002874cachelets for normal memory accesses. The semantics of the Alpha removes the
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002875need for coordination in the absence of memory barriers.
David Howells108b42b2006-03-31 16:00:29 +01002876
2877
2878CACHE COHERENCY VS DMA
2879----------------------
2880
2881Not all systems maintain cache coherency with respect to devices doing DMA. In
2882such cases, a device attempting DMA may obtain stale data from RAM because
2883dirty cache lines may be resident in the caches of various CPUs, and may not
2884have been written back to RAM yet. To deal with this, the appropriate part of
2885the kernel must flush the overlapping bits of cache on each CPU (and maybe
2886invalidate them as well).
2887
2888In addition, the data DMA'd to RAM by a device may be overwritten by dirty
2889cache lines being written back to RAM from a CPU's cache after the device has
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002890installed its own data, or cache lines present in the CPU's cache may simply
2891obscure the fact that RAM has been updated, until at such time as the cacheline
2892is discarded from the CPU's cache and reloaded. To deal with this, the
2893appropriate part of the kernel must invalidate the overlapping bits of the
David Howells108b42b2006-03-31 16:00:29 +01002894cache on each CPU.
2895
2896See Documentation/cachetlb.txt for more information on cache management.
2897
2898
2899CACHE COHERENCY VS MMIO
2900-----------------------
2901
2902Memory mapped I/O usually takes place through memory locations that are part of
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002903a window in the CPU's memory space that has different properties assigned than
David Howells108b42b2006-03-31 16:00:29 +01002904the usual RAM directed window.
2905
2906Amongst these properties is usually the fact that such accesses bypass the
2907caching entirely and go directly to the device buses. This means MMIO accesses
2908may, in effect, overtake accesses to cached memory that were emitted earlier.
2909A memory barrier isn't sufficient in such a case, but rather the cache must be
2910flushed between the cached memory write and the MMIO access if the two are in
2911any way dependent.
2912
2913
2914=========================
2915THE THINGS CPUS GET UP TO
2916=========================
2917
2918A programmer might take it for granted that the CPU will perform memory
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002919operations in exactly the order specified, so that if the CPU is, for example,
David Howells108b42b2006-03-31 16:00:29 +01002920given the following piece of code to execute:
2921
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002922 a = READ_ONCE(*A);
2923 WRITE_ONCE(*B, b);
2924 c = READ_ONCE(*C);
2925 d = READ_ONCE(*D);
2926 WRITE_ONCE(*E, e);
David Howells108b42b2006-03-31 16:00:29 +01002927
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002928they would then expect that the CPU will complete the memory operation for each
David Howells108b42b2006-03-31 16:00:29 +01002929instruction before moving on to the next one, leading to a definite sequence of
2930operations as seen by external observers in the system:
2931
2932 LOAD *A, STORE *B, LOAD *C, LOAD *D, STORE *E.
2933
2934
2935Reality is, of course, much messier. With many CPUs and compilers, the above
2936assumption doesn't hold because:
2937
2938 (*) loads are more likely to need to be completed immediately to permit
2939 execution progress, whereas stores can often be deferred without a
2940 problem;
2941
2942 (*) loads may be done speculatively, and the result discarded should it prove
2943 to have been unnecessary;
2944
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002945 (*) loads may be done speculatively, leading to the result having been fetched
2946 at the wrong time in the expected sequence of events;
David Howells108b42b2006-03-31 16:00:29 +01002947
2948 (*) the order of the memory accesses may be rearranged to promote better use
2949 of the CPU buses and caches;
2950
2951 (*) loads and stores may be combined to improve performance when talking to
2952 memory or I/O hardware that can do batched accesses of adjacent locations,
2953 thus cutting down on transaction setup costs (memory and PCI devices may
2954 both be able to do this); and
2955
2956 (*) the CPU's data cache may affect the ordering, and whilst cache-coherency
2957 mechanisms may alleviate this - once the store has actually hit the cache
2958 - there's no guarantee that the coherency management will be propagated in
2959 order to other CPUs.
2960
2961So what another CPU, say, might actually observe from the above piece of code
2962is:
2963
2964 LOAD *A, ..., LOAD {*C,*D}, STORE *E, STORE *B
2965
2966 (Where "LOAD {*C,*D}" is a combined load)
2967
2968
2969However, it is guaranteed that a CPU will be self-consistent: it will see its
2970_own_ accesses appear to be correctly ordered, without the need for a memory
2971barrier. For instance with the following code:
2972
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002973 U = READ_ONCE(*A);
2974 WRITE_ONCE(*A, V);
2975 WRITE_ONCE(*A, W);
2976 X = READ_ONCE(*A);
2977 WRITE_ONCE(*A, Y);
2978 Z = READ_ONCE(*A);
David Howells108b42b2006-03-31 16:00:29 +01002979
2980and assuming no intervention by an external influence, it can be assumed that
2981the final result will appear to be:
2982
2983 U == the original value of *A
2984 X == W
2985 Z == Y
2986 *A == Y
2987
2988The code above may cause the CPU to generate the full sequence of memory
2989accesses:
2990
2991 U=LOAD *A, STORE *A=V, STORE *A=W, X=LOAD *A, STORE *A=Y, Z=LOAD *A
2992
2993in that order, but, without intervention, the sequence may have almost any
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002994combination of elements combined or discarded, provided the program's view
2995of the world remains consistent. Note that READ_ONCE() and WRITE_ONCE()
2996are -not- optional in the above example, as there are architectures
2997where a given CPU might reorder successive loads to the same location.
2998On such architectures, READ_ONCE() and WRITE_ONCE() do whatever is
2999necessary to prevent this, for example, on Itanium the volatile casts
3000used by READ_ONCE() and WRITE_ONCE() cause GCC to emit the special ld.acq
3001and st.rel instructions (respectively) that prevent such reordering.
David Howells108b42b2006-03-31 16:00:29 +01003002
3003The compiler may also combine, discard or defer elements of the sequence before
3004the CPU even sees them.
3005
3006For instance:
3007
3008 *A = V;
3009 *A = W;
3010
3011may be reduced to:
3012
3013 *A = W;
3014
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003015since, without either a write barrier or an WRITE_ONCE(), it can be
Paul E. McKenney2ecf8102013-12-11 13:59:04 -08003016assumed that the effect of the storage of V to *A is lost. Similarly:
David Howells108b42b2006-03-31 16:00:29 +01003017
3018 *A = Y;
3019 Z = *A;
3020
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003021may, without a memory barrier or an READ_ONCE() and WRITE_ONCE(), be
3022reduced to:
David Howells108b42b2006-03-31 16:00:29 +01003023
3024 *A = Y;
3025 Z = Y;
3026
3027and the LOAD operation never appear outside of the CPU.
3028
3029
3030AND THEN THERE'S THE ALPHA
3031--------------------------
3032
3033The DEC Alpha CPU is one of the most relaxed CPUs there is. Not only that,
3034some versions of the Alpha CPU have a split data cache, permitting them to have
Jarek Poplawski81fc6322007-05-23 13:58:20 -07003035two semantically-related cache lines updated at separate times. This is where
David Howells108b42b2006-03-31 16:00:29 +01003036the data dependency barrier really becomes necessary as this synchronises both
3037caches with the memory coherence system, thus making it seem like pointer
3038changes vs new data occur in the right order.
3039
Jarek Poplawski81fc6322007-05-23 13:58:20 -07003040The Alpha defines the Linux kernel's memory barrier model.
David Howells108b42b2006-03-31 16:00:29 +01003041
3042See the subsection on "Cache Coherency" above.
3043
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003044VIRTUAL MACHINE GUESTS
3045-------------------
3046
3047Guests running within virtual machines might be affected by SMP effects even if
3048the guest itself is compiled without SMP support. This is an artifact of
3049interfacing with an SMP host while running an UP kernel. Using mandatory
3050barriers for this use-case would be possible but is often suboptimal.
3051
3052To handle this case optimally, low-level virt_mb() etc macros are available.
3053These have the same effect as smp_mb() etc when SMP is enabled, but generate
3054identical code for SMP and non-SMP systems. For example, virtual machine guests
3055should use virt_mb() rather than smp_mb() when synchronizing against a
3056(possibly SMP) host.
3057
3058These are equivalent to smp_mb() etc counterparts in all other respects,
3059in particular, they do not control MMIO effects: to control
3060MMIO effects, use mandatory barriers.
David Howells108b42b2006-03-31 16:00:29 +01003061
David Howells90fddab2010-03-24 09:43:00 +00003062============
3063EXAMPLE USES
3064============
3065
3066CIRCULAR BUFFERS
3067----------------
3068
3069Memory barriers can be used to implement circular buffering without the need
3070of a lock to serialise the producer with the consumer. See:
3071
3072 Documentation/circular-buffers.txt
3073
3074for details.
3075
3076
David Howells108b42b2006-03-31 16:00:29 +01003077==========
3078REFERENCES
3079==========
3080
3081Alpha AXP Architecture Reference Manual, Second Edition (Sites & Witek,
3082Digital Press)
3083 Chapter 5.2: Physical Address Space Characteristics
3084 Chapter 5.4: Caches and Write Buffers
3085 Chapter 5.5: Data Sharing
3086 Chapter 5.6: Read/Write Ordering
3087
3088AMD64 Architecture Programmer's Manual Volume 2: System Programming
3089 Chapter 7.1: Memory-Access Ordering
3090 Chapter 7.4: Buffering and Combining Memory Writes
3091
3092IA-32 Intel Architecture Software Developer's Manual, Volume 3:
3093System Programming Guide
3094 Chapter 7.1: Locked Atomic Operations
3095 Chapter 7.2: Memory Ordering
3096 Chapter 7.4: Serializing Instructions
3097
3098The SPARC Architecture Manual, Version 9
3099 Chapter 8: Memory Models
3100 Appendix D: Formal Specification of the Memory Models
3101 Appendix J: Programming with the Memory Models
3102
3103UltraSPARC Programmer Reference Manual
3104 Chapter 5: Memory Accesses and Cacheability
3105 Chapter 15: Sparc-V9 Memory Models
3106
3107UltraSPARC III Cu User's Manual
3108 Chapter 9: Memory Models
3109
3110UltraSPARC IIIi Processor User's Manual
3111 Chapter 8: Memory Models
3112
3113UltraSPARC Architecture 2005
3114 Chapter 9: Memory
3115 Appendix D: Formal Specifications of the Memory Models
3116
3117UltraSPARC T1 Supplement to the UltraSPARC Architecture 2005
3118 Chapter 8: Memory Models
3119 Appendix F: Caches and Cache Coherency
3120
3121Solaris Internals, Core Kernel Architecture, p63-68:
3122 Chapter 3.3: Hardware Considerations for Locks and
3123 Synchronization
3124
3125Unix Systems for Modern Architectures, Symmetric Multiprocessing and Caching
3126for Kernel Programmers:
3127 Chapter 13: Other Memory Models
3128
3129Intel Itanium Architecture Software Developer's Manual: Volume 1:
3130 Section 2.6: Speculation
3131 Section 4.4: Memory Access